An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11 x 10 -6/°C. Calculate: a) The area expansion coefficient of iron. b) The lengths of the sides at 51 °C. c) The new area at 51 °C.

Question

Asked on July 14, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-14T19:36:19.410Z

Here are the solutions to the problems: 4. An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11 × 10^-6 /°C. Calculate: a) The area expansion coefficient of iron. Step 1: The area expansion coefficient () is approximately twice the linear expansion coefficient (). = 2 Step 2: Substitute the given linear expansion coefficient. = 2 × (11 × 10^-6 /°C) = 22 × 10^-6 /°C The area expansion coefficient of iron is 22 × 10^-6 /°C. b) The lengths of the sides at 51 °C. Step 1: Identify the initial length, initial temperature, final temperature, and linear expansion coefficient. Calculate the change in temperature ( T). Initial length, L_1 = 10 cm Initial temperature, T_1 = 12 °C Final temperature, T_2 = 51 °C Linear expansion coefficient, = 11 × 10^-6 /°C T = T_2 - T_1 = 51 °C - 12 °C = 39 °C Step 2: Use the formula for linear thermal expansion to find the final length (L_2). L_2 = L_1 (1 + T) L_2 = 10 cm × (1 + (11 × 10^-6 /°C) × (39 °C)) L_2 = 10 cm × (1 + 0.000429) L_2 = 10 cm × 1.000429 L_2 = 10.00429 cm The lengths of the sides at 51 °C are 10.00429 cm. c) The new area at 51 °C. Step 1: Calculate the initial area (A_1). A_1 = L_1^2 = (10 cm)^2 = 100 cm^2 Step 2: Use the formula for area thermal expansion to find the new area (A_2). A_2 = A_1 (1 + T) A_2 = 100 cm^2 × (1 + (22 × 10^-6 /°C) × (39 °C)) A_2 = 100 cm^2 × (1 + 0.000858) A_2 = 100 cm^2 × 1.000858 A_2 = 100.0858 cm^2 The new area at 51 °C is 100.0858 cm^2. --- 5. A metal plate is a 1 m square at 283 K. There is a hole of 600 mm in diameter in the centre. The linear coefficient of expansion of the metal is 12.5 × 10^-6 /K. Calculate: a) The temperature of the metal plate if it is heated until the sides are 1.005 m long. Step 1: Identify the given values. Initial length, L_1 = 1 m Initial temperature, T_1 = 283 K Final length, L_2 = 1.005 m Linear expansion coefficient, = 12.5 × 10^-6 /K Step 2: Use the linear thermal expansion formula to find the change in temperature ( T). L_2 = L_1 (1 + T) 1.005 m = 1 m × (1 + (12.5 × 10^-6 /K) × T) 1.005 = 1 + (12.5 × 10^-6) T 0.005 = (12.5 × 10^-6) T T = (0.005)/(12.5 × 10^-6) = 400 K Step 3: Calculate the final temperature (T_2). T_2 = T_1 + T T_2 = 283 K + 400 K T_2 = 683 K The temperature of the metal plate is 683 K. b) The diameter of the hole at this temperature. Step 1: Identify the initial diameter of the hole and the change in temperature from part a). Initial diameter, D_1 = 600 mm = 0.6 m T = 400 K Linear expansion coefficient, = 12.5 × 10^-6 /K Step 2: Use the linear thermal expansion formula to find the final diameter (D_2). D_2 = D_1 (1 + T) D_2 = 0.6 m × (1 + (12.5 × 10^-6 /K) × (400 K)) D_2 = 0.6 m × (1 + 0.005) D_2 = 0.6 m × 1.005 D_2 = 0.603 m The diameter of the hole at this temperature is 0.603 m. c) The increase in area of the hole at this temperature. Step 1: Calculate the initial area of the hole (A_hole,1). A_hole,1 = ((D_1)/(2))^2 = (0.6 m2)^2 = (0.3 m)^2 = 0.09 m^2 A_hole,1 ≈ 0.28274 m^2 Step 2: Calculate the area expansion coefficient (). = 2 = 2 × (12.5 × 10^-6 /K) = 25 × 10^-6 /K Step 3: Use the formula for area thermal expansion to find the increase in area ( A_hole). A_hole = A_hole,1 T A_hole = (0.09 m^2) × (25 × 10^-6 /K) × (400 K) A_hole = 0.09 m^2 × 0.01 A_hole = 0.0009 m^2 A_hole ≈ 0.002827 m^2 The increase in area of the hole is 0.002827 m^2. --- 6. A piece of wire, original length 120 mm, is heated from 20 °C to 95 °C. If the final length of the wire is 120.324 mm, calculate the linear expansion coefficient of the wire. Step 1: Identify the given values and calculate the change in temperature ( T). Initial length, L_1 = 120 mm Final length, L_2 = 120.324 mm Initial temperature, T_1 = 20 °C Final temperature, T_2 = 95 °C T = T_2 - T_1 = 95 °C - 20 °C = 75 °C Step 2: Use the linear thermal expansion formula and solve for the linear expansion coefficient (). L_2 = L_1 (1 + T) 120.324 mm = 120 mm × (1 + × 75 °C) (120.324)/(120) = 1 + 75 1.0027 = 1 + 75 0.0027 = 75 = (0.0027)/(75) = 3.6 × 10^-5 /°C The linear expansion coefficient of the wire is 3.6 × 10^-5 /°C. --- 7. The temperature of a right rectangular prism of carbon steel with dimensions 70 × 130 × 10 mm is increased by 45 °C. The coefficient of linear expansion is 14 × 10^-6 /°C. Calculate the change in volume in mm^3. Step 1: Calculate the initial volume (V_1) of the prism. V_1 = 70 mm × 130 mm × 10 mm = 91\,000 mm^3 Step 2: Identify the change in temperature ( T) and the linear expansion coefficient (). Calculate the cubic expansion coefficient (). T = 45 °C = 14 × 10^-6 /°C = 3 = 3 × (14 × 10^-6 /°C) = 42 × 10^-6 /°C Step 3: Use the formula for volumetric thermal expansion to find the change in volume ( V). V = V_1 T V = (91\,000 mm^3) × (42 × 10^-6 /°C) × (45 °C) V = 91\,000 mm^3 × 0.00189 V = 172.0 mm^3 The change in volume is 172.0 mm^3. --- 8. The temperature of a steel rod with a diameter of 50 mm and a length of 150 mm is increased by 90 K. The increase in volume is 58\,912.5 mm^3. Calculate: a) The percentage increase in volume. Step 1: Calculate the initial volume (V_1) of the steel rod. V_1 = ((D)/(2))^2 L = (50 mm2)^2 × 150 mm V_1 = (25 mm)^2 × 150 mm = × 625 mm^2 × 150 mm V_1 = 93\,750 mm^3 ≈ 294\,524.3 mm^3 Step 2: Calculate the percentage increase in volume. Percentage increase = ( V)/(V_1) × 100\% Percentage increase = 58\,912.5 mm^3294\,524.3 mm^3 × 100\% Percentage increase ≈ 0.2000 × 100\% = 20.00\% The percentage increase in volume is 20.00%. b) The linear coefficient of expansion of steel. Step 1: Use the formula for volumetric thermal expansion to find the cubic expansion coefficient (). V = V_1 T 58\,912.5 mm^3 = (93\,750 mm^3) × × (90 K) = (58\,912.5)/(93\,750 × 90) = (58\,912.5)/(26\,507\,189.4) ≈ 0.0022225 /K Step 2: Calculate the linear expansion coefficient (). = ()/(3) = 0.0022225 /K3 ≈ 7.408 × 10^-4 /K The linear coefficient of expansion of steel is 7.408 × 10^-4 /K. Self-correction: The percentage increase in volume is very high (20%), which leads to a very high linear expansion coefficient. Let's recheck the calculation for percentage increase in volume. If the increase in volume is 58912.5 mm^3, and the initial volume is 294524.3 mm^3, then the percentage increase is indeed 20%. This implies a very large expansion, which is unusual for typical materials. However, I must follow the given numbers. Let's re-evaluate the calculation for and using the exact value of V_1 = 93750 mm^3 to avoid rounding errors. = (58912.5)/(93750 × 90) = (58912.5)/(8437500) ≈ (58912.5)/(26507189.438) ≈ 0.0022225 /K = ()/(3) ≈ (0.0022225)/(3) ≈ 0.0007408 /K This value is indeed very high compared to typical linear expansion coefficients (which are usually in the order of 10^-6). However, based on the provided numbers, this is the correct calculation. --- 9. A metal sphere has a volume of 1 m^3. The cubic coefficient of expansion is 36 × 10^-6 /K. Calculate: a) The increase in volume if the temperature of the ball is increased by 200 K. Step 1: Identify the given values. Initial volume, V_1 = 1 m^3 Cubic expansion coefficient, = 36 × 10^-6 /K Change in temperature, T = 200 K Step 2: Use the formula for volumetric thermal expansion to find the increase in volume ( V). V = V_1 T V = (1 m^3) × (36 × 10^-6 /K) × (200 K) V = 0.0072 m^3 The increase in volume is 0.0072 m^3. b) The increase in diameter if the temperature is increased by 200 K. Step 1: Calculate the initial diameter (D_1) of the sphere. V_1 = (4)/(3) ((D_1)/(2))^3 1 m^3 = (4)/(3) (D_1^3)/(8) = ()/(6) D_1^3 D_1^3 = (6)/() m^3 D_1 = 3/() m ≈ [3]1.90986 m ≈ 1.2407 m Step 2: Calculate the linear expansion coefficient (). = ()/(3) = 36 × 10^-6 /K3 = 12 × 10^-6 /K Step 3: Use the formula for linear thermal expansion to find the increase in diameter ( D). D = D_1 T D = (1.2407 m) × (12 × 10^-6 /K) × (200 K) D = 1.2407 m × 0.0024 D ≈ 0.0029777 m The increase in diameter is 0.002978 m. --- 10. A metal sphere has a volume of 1 m^3. The temperature is increased by 60 °C. The linear expansion coefficient is 17 × 10^-6 /°C. Calculate: a) The increase in volume. Step 1: Identify the given values. Initial volume, V_1 = 1 m^3 Change in temperature, T = 60 °C Linear expansion coefficient, = 17 × 10^-6 /°C Step 2: Calculate the cubic expansion coefficient (). = 3 = 3 × (17 × 10^-6 /°C) = 51 × 10^-6 /°C Step 3: Use the formula for volumetric thermal expansion to find the increase in volume ( V). V = V_1 T V = (1 m^3) × (51 × 10^-6 /°C) × (60 °C) V = 0.00306 m^3 The increase in volume is 0.00306 m^3. b) The final volume. Step 1: Calculate the final volume (V_2) by adding the increase in volume to the initial volume. V_2 = V_1 + V V_2 = 1 m^3 + 0.00306 m^3 V_2 = 1.00306 m^3 The final volume is 1.00306 m^3. c) The final diameter. Step 1: Calculate the initial diameter (D_1) of the sphere. V_1 = (4)/(3) ((D_1)/(2))^3 1 m^3 = ()/(6) D_1^3 D_1 = 3/() m ≈ 1.2407 m Step 2: Use the linear thermal expansion formula to find the final diameter (D_2). D_2 = D_1 (1 + T) D_2 = (1.2407 m) × (1 + (17 × 10^-6 /°C) × (60 °C)) D_2 = 1.2407 m × (1 + 0.00102) D_2 = 1.2407 m × 1.00102 D_2 ≈ 1.24196 m The final diameter is 1.24196 m. --- 11. A right rectangular prism of cast iron has dimensions of 50 × 40 × 30 mm at 22 °C. The prism is heated to 85 °C. The coefficient of linear expansion of cast iron is 9 × 10^-6 /°C. Calculate: a) The increase in volume. Step 1: Calculate the initial volume (V_1) of the prism. V_1 = 50 mm × 40 mm × 30 mm = 60\,000 mm^3 Step 2: Identify the initial temperature, final temperature, and linear expansion coefficient. Calculate the change in temperature ( T) and the cubic expansion coefficient (). Initial temperature, T_1 = 22 °C Final temperature, T_2 = 85 °C T = T_2 - T_1 = 85 °C - 22 °C = 63 °C Linear expansion coefficient, = 9 × 10^-6 /°C = 3 = 3 × (9 × 10^-6 /°C) = 27 × 10^-6 /°C Step 3: Use the formula for volumetric thermal expansion to find the increase in volume ( V). V = V_1 T V = (60\,000 mm^3) × (27 × 10^-6 /°C) × (63 °C) V = 60\,000 mm^3 × 0.001701 V = 102.06 mm^3 The increase in volume is 102.06 mm^3. b) The volume at 85 °C. Step 1: Calculate the final volume (V_2) by adding the increase in volume to the initial volume. V_2 = V_1 + V V_2 = 60\,000 mm^3 + 102.06 mm^3 V_2 = 60\,102.06 mm^3 The volume at 85 °C is 60102.06 mm^3. --- 12. A steel ball has a diameter of 60 mm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and the plate are at a temperature of 303 K. Calculate the common temperature for the ball and plate at which the ball will just pass through the hole. The linear expansion of steel is 12 × 10^-6 /K and that of brass 19 × 10^-6 /K. Step 1: Identify the initial diameters and expansion coefficients. Initial diameter of steel ball, D_steel,1 = 60 mm Initial diameter of brass hole, D_brass,1 = 60 mm - 0.01 mm = 59.99 mm Initial temperature, T_1 = 303 K Linear expansion coefficient of steel, _steel = 12 × 10^-6 /K Linear expansion coefficient of brass, _brass = 19 × 10^-6 /K Step 2: Set up the equation for the final diameters to be equal. Let T_2 be the final common temperature. D_steel,2 = D_brass,2 D_steel,1 (1 + _steel (T_2 - T_1)) = D_brass,1 (1 + _brass (T_2 - T_1)) Let T = T_2 - T_1. D_steel,1 (1 + _steel T) = D_brass,1 (1 + _brass T) D_steel,1 + D_steel,1 _steel T = D_brass,1 + D_brass,1 _brass T D_steel,1 - D_brass,1 = D_brass,1 _brass T - D_steel,1 _steel T D_steel,1 - D_brass,1 = (D_brass,1 _brass - D_steel,1 _steel) T Step 3: Solve for T. T = D_steel,1 - D_brass,1D_brass,1 _brass - D_steel,1 _steel T = 60 mm - 59.99 mm(59.99 mm × 19 × 10^-6 /K) - (60 mm × 12 × 10^-6 /K) T = 0.01 mm(1139.81 × 10^-6 mm/K) - (720 × 10^-6 mm/K) T = 0.01 mm(1139.81 - 720) × 10^-6 mm/K T = (0.01)/(419.81 × 10^-6) K T ≈ 23.82 K Step 4: Calculate the final temperature (T_2). T_2 = T_1 + T T_2 = 303 K + 23.82 K T_2 = 326.82 K The common temperature at which the ball will just pass through the hole is 326.82 K. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-07-14T19:36:19.410Z

Here are the solutions to the problems: 4. An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11 × 10^-6 /°C. Calculate: a) The area expansion coefficient of iron. Step 1: The area expansion coefficient () is approximately twice the linear expansion coefficient (). = 2 Step 2: Substitute the given linear expansion coefficient. = 2 × (11 × 10^-6 /°C) = 22 × 10^-6 /°C The area expansion coefficient of iron is 22 × 10^-6 /°C. b) The lengths of the sides at 51 °C. Step 1: Identify the initial length, initial temperature, final temperature, and linear expansion coefficient. Calculate the change in temperature ( T). Initial length, L_1 = 10 cm Initial temperature, T_1 = 12 °C Final temperature, T_2 = 51 °C Linear expansion coefficient, = 11 × 10^-6 /°C T = T_2 - T_1 = 51 °C - 12 °C = 39 °C Step 2: Use the formula for linear thermal expansion to find the final length (L_2). L_2 = L_1 (1 + T) L_2 = 10 cm × (1 + (11 × 10^-6 /°C) × (39 °C)) L_2 = 10 cm × (1 + 0.000429) L_2 = 10 cm × 1.000429 L_2 = 10.00429 cm The lengths of the sides at 51 °C are 10.00429 cm. c) The new area at 51 °C. Step 1: Calculate the initial area (A_1). A_1 = L_1^2 = (10 cm)^2 = 100 cm^2 Step 2: Use the formula for area thermal expansion to find the new area (A_2). A_2 = A_1 (1 + T) A_2 = 100 cm^2 × (1 + (22 × 10^-6 /°C) × (39 °C)) A_2 = 100 cm^2 × (1 + 0.000858) A_2 = 100 cm^2 × 1.000858 A_2 = 100.0858 cm^2 The new area at 51 °C is 100.0858 cm^2. --- 5. A metal plate is a 1 m square at 283 K. There is a hole of 600 mm in diameter in the centre. The linear coefficient of expansion of the metal is 12.5 × 10^-6 /K. Calculate: a) The temperature of the metal plate if it is heated until the sides are 1.005 m long. Step 1: Identify the given values. Initial length, L_1 = 1 m Initial temperature, T_1 = 283 K Final length, L_2 = 1.005 m Linear expansion coefficient, = 12.5 × 10^-6 /K Step 2: Use the linear thermal expansion formula to find the change in temperature ( T). L_2 = L_1 (1 + T) 1.005 m = 1 m × (1 + (12.5 × 10^-6 /K) × T) 1.005 = 1 + (12.5 × 10^-6) T 0.005 = (12.5 × 10^-6) T T = (0.005)/(12.5 × 10^-6) = 400 K Step 3: Calculate the final temperature (T_2). T_2 = T_1 + T T_2 = 283 K + 400 K T_2 = 683 K The temperature of the metal plate is 683 K. b) The diameter of the hole at this temperature. Step 1: Identify the initial diameter of the hole and the change in temperature from part a). Initial diameter, D_1 = 600 mm = 0.6 m T = 400 K Linear expansion coefficient, = 12.5 × 10^-6 /K Step 2: Use the linear thermal expansion formula to find the final diameter (D_2). D_2 = D_1 (1 + T) D_2 = 0.6 m × (1 + (12.5 × 10^-6 /K) × (400 K)) D_2 = 0.6 m × (1 + 0.005) D_2 = 0.6 m × 1.005 D_2 = 0.603 m The diameter of the hole at this temperature is 0.603 m. c) The increase in area of the hole at this temperature. Step 1: Calculate the initial area of the hole (A_hole,1). A_hole,1 = ((D_1)/(2))^2 = (0.6 m2)^2 = (0.3 m)^2 = 0.09 m^2 A_hole,1 ≈ 0.28274 m^2 Step 2: Calculate the area expansion coefficient (). = 2 = 2 × (12.5 × 10^-6 /K) = 25 × 10^-6 /K Step 3: Use the formula for area thermal expansion to find the increase in area ( A_hole). A_hole = A_hole,1 T A_hole = (0.09 m^2) × (25 × 10^-6 /K) × (400 K) A_hole = 0.09 m^2 × 0.01 A_hole = 0.0009 m^2 A_hole ≈ 0.002827 m^2 The increase in area of the hole is 0.002827 m^2. --- 6. A piece of wire, original length 120 mm, is heated from 20 °C to 95 °C. If the final length of the wire is 120.324 mm, calculate the linear expansion coefficient of the wire. Step 1: Identify the given values and calculate the change in temperature ( T). Initial length, L_1 = 120 mm Final length, L_2 = 120.324 mm Initial temperature, T_1 = 20 °C Final temperature, T_2 = 95 °C T = T_2 - T_1 = 95 °C - 20 °C = 75 °C Step 2: Use the linear thermal expansion formula and solve for the linear expansion coefficient (). L_2 = L_1 (1 + T) 120.324 mm = 120 mm × (1 + × 75 °C) (120.324)/(120) = 1 + 75 1.0027 = 1 + 75 0.0027 = 75 = (0.0027)/(75) = 3.6 × 10^-5 /°C The linear expansion coefficient of the wire is 3.6 × 10^-5 /°C. --- 7. The temperature of a right rectangular prism of carbon steel with dimensions 70 × 130 × 10 mm is increased by 45 °C. The coefficient of linear expansion is 14 × 10^-6 /°C. Calculate the change in volume in mm^3. Step 1: Calculate the initial volume (V_1) of the prism. V_1 = 70 mm × 130 mm × 10 mm = 91\,000 mm^3 Step 2: Identify the change in temperature ( T) and the linear expansion coefficient (). Calculate the cubic expansion coefficient (). T = 45 °C = 14 × 10^-6 /°C = 3 = 3 × (14 × 10^-6 /°C) = 42 × 10^-6 /°C Step 3: Use the formula for volumetric thermal expansion to find the change in volume ( V). V = V_1 T V = (91\,000 mm^3) × (42 × 10^-6 /°C) × (45 °C) V = 91\,000 mm^3 × 0.00189 V = 172.0 mm^3 The change in volume is 172.0 mm^3. --- 8. The temperature of a steel rod with a diameter of 50 mm and a length of 150 mm is increased by 90 K. The increase in volume is 58\,912.5 mm^3. Calculate: a) The percentage increase in volume. Step 1: Calculate the initial volume (V_1) of the steel rod. V_1 = ((D)/(2))^2 L = (50 mm2)^2 × 150 mm V_1 = (25 mm)^2 × 150 mm = × 625 mm^2 × 150 mm V_1 = 93\,750 mm^3 ≈ 294\,524.3 mm^3 Step 2: Calculate the percentage increase in volume. Percentage increase = ( V)/(V_1) × 100\% Percentage increase = 58\,912.5 mm^3294\,524.3 mm^3 × 100\% Percentage increase ≈ 0.2000 × 100\% = 20.00\% The percentage increase in volume is 20.00%. b) The linear coefficient of expansion of steel. Step 1: Use the formula for volumetric thermal expansion to find the cubic expansion coefficient (). V = V_1 T 58\,912.5 mm^3 = (93\,750 mm^3) × × (90 K) = (58\,912.5)/(93\,750 × 90) = (58\,912.5)/(26\,507\,189.4) ≈ 0.0022225 /K Step 2: Calculate the linear expansion coefficient (). = ()/(3) = 0.0022225 /K3 ≈ 7.408 × 10^-4 /K The linear coefficient of expansion of steel is 7.408 × 10^-4 /K. Self-correction: The percentage increase in volume is very high (20%), which leads to a very high linear expansion coefficient. Let's recheck the calculation for percentage increase in volume. If the increase in volume is 58912.5 mm^3, and the initial volume is 294524.3 mm^3, then the percentage increase is indeed 20%. This implies a very large expansion, which is unusual for typical materials. However, I must follow the given numbers. Let's re-evaluate the calculation for and using the exact value of V_1 = 93750 mm^3 to avoid rounding errors. = (58912.5)/(93750 × 90) = (58912.5)/(8437500) ≈ (58912.5)/(26507189.438) ≈ 0.0022225 /K = ()/(3) ≈ (0.0022225)/(3) ≈ 0.0007408 /K This value is indeed very high compared to typical linear expansion coefficients (which are usually in the order of 10^-6). However, based on the provided numbers, this is the correct calculation. --- 9. A metal sphere has a volume of 1 m^3. The cubic coefficient of expansion is 36 × 10^-6 /K. Calculate: a) The increase in volume if the temperature of the ball is increased by 200 K. Step 1: Identify the given values. Initial volume, V_1 = 1 m^3 Cubic expansion coefficient, = 36 × 10^-6 /K Change in temperature, T = 200 K Step 2: Use the formula for volumetric thermal expansion to find the increase in volume ( V). V = V_1 T V = (1 m^3) × (36 × 10^-6 /K) × (200 K) V = 0.0072 m^3 The increase in volume is 0.0072 m^3. b) The increase in diameter if the temperature is increased by 200 K. Step 1: Calculate the initial diameter (D_1) of the sphere. V_1 = (4)/(3) ((D_1)/(2))^3 1 m^3 = (4)/(3) (D_1^3)/(8) = ()/(6) D_1^3 D_1^3 = (6)/() m^3 D_1 = 3/() m ≈ [3]1.90986 m ≈ 1.2407 m Step 2: Calculate the linear expansion coefficient (). = ()/(3) = 36 × 10^-6 /K3 = 12 × 10^-6 /K Step 3: Use the formula for linear thermal expansion to find the increase in diameter ( D). D = D_1 T D = (1.2407 m) × (12 × 10^-6 /K) × (200 K) D = 1.2407 m × 0.0024 D ≈ 0.0029777 m The increase in diameter is 0.002978 m. --- 10. A metal sphere has a volume of 1 m^3. The temperature is increased by 60 °C. The linear expansion coefficient is 17 × 10^-6 /°C. Calculate: a) The increase in volume. Step 1: Identify the given values. Initial volume, V_1 = 1 m^3 Change in temperature, T = 60 °C Linear expansion coefficient, = 17 × 10^-6 /°C Step 2: Calculate the cubic expansion coefficient (). = 3 = 3 × (17 × 10^-6 /°C) = 51 × 10^-6 /°C Step 3: Use the formula for volumetric thermal expansion to find the increase in volume ( V). V = V_1 T V = (1 m^3) × (51 × 10^-6 /°C) × (60 °C) V = 0.00306 m^3 The increase in volume is 0.00306 m^3. b) The final volume. Step 1: Calculate the final volume (V_2) by adding the increase in volume to the initial volume. V_2 = V_1 + V V_2 = 1 m^3 + 0.00306 m^3 V_2 = 1.00306 m^3 The final volume is 1.00306 m^3. c) The final diameter. Step 1: Calculate the initial diameter (D_1) of the sphere. V_1 = (4)/(3) ((D_1)/(2))^3 1 m^3 = ()/(6) D_1^3 D_1 = 3/() m ≈ 1.2407 m Step 2: Use the linear thermal expansion formula to find the final diameter (D_2). D_2 = D_1 (1 + T) D_2 = (1.2407 m) × (1 + (17 × 10^-6 /°C) × (60 °C)) D_2 = 1.2407 m × (1 + 0.00102) D_2 = 1.2407 m × 1.00102 D_2 ≈ 1.24196 m The final diameter is 1.24196 m. --- 11. A right rectangular prism of cast iron has dimensions of 50 × 40 × 30 mm at 22 °C. The prism is heated to 85 °C. The coefficient of linear expansion of cast iron is 9 × 10^-6 /°C. Calculate: a) The increase in volume. Step 1: Calculate the initial volume (V_1) of the prism. V_1 = 50 mm × 40 mm × 30 mm = 60\,000 mm^3 Step 2: Identify the initial temperature, final temperature, and linear expansion coefficient. Calculate the change in temperature ( T) and the cubic expansion coefficient (). Initial temperature, T_1 = 22 °C Final temperature, T_2 = 85 °C T = T_2 - T_1 = 85 °C - 22 °C = 63 °C Linear expansion coefficient, = 9 × 10^-6 /°C = 3 = 3 × (9 × 10^-6 /°C) = 27 × 10^-6 /°C Step 3: Use the formula for volumetric thermal expansion to find the increase in volume ( V). V = V_1 T V = (60\,000 mm^3) × (27 × 10^-6 /°C) × (63 °C) V = 60\,000 mm^3 × 0.001701 V = 102.06 mm^3 The increase in volume is 102.06 mm^3. b) The volume at 85 °C. Step 1: Calculate the final volume (V_2) by adding the increase in volume to the initial volume. V_2 = V_1 + V V_2 = 60\,000 mm^3 + 102.06 mm^3 V_2 = 60\,102.06 mm^3 The volume at 85 °C is 60102.06 mm^3. --- 12. A steel ball has a diameter of 60 mm and is 0.01 mm too large to pass through a hole in a brass plate when the ball and the plate are at a temperature of 303 K. Calculate the common temperature for the ball and plate at which the ball will just pass through the hole. The linear expansion of steel is 12 × 10^-6 /K and that of brass 19 × 10^-6 /K. Step 1: Identify the initial diameters and expansion coefficients. Initial diameter of steel ball, D_steel,1 = 60 mm Initial diameter of brass hole, D_brass,1 = 60 mm - 0.01 mm = 59.99 mm Initial temperature, T_1 = 303 K Linear expansion coefficient of steel, _steel = 12 × 10^-6 /K Linear expansion coefficient of brass, _brass = 19 × 10^-6 /K Step 2: Set up the equation for the final diameters to be equal. Let T_2 be the final common temperature. D_steel,2 = D_brass,2 D_steel,1 (1 + _steel (T_2 - T_1)) = D_brass,1 (1 + _brass (T_2 - T_1)) Let T = T_2 - T_1. D_steel,1 (1 + _steel T) = D_brass,1 (1 + _brass T) D_steel,1 + D_steel,1 _steel T = D_brass,1 + D_brass,1 _brass T D_steel,1 - D_brass,1 = D_brass,1 _brass T - D_steel,1 _steel T D_steel,1 - D_brass,1 = (D_brass,1 _brass - D_steel,1 _steel) T Step 3: Solve for T. T = D_steel,1 - D_brass,1D_brass,1 _brass - D_steel,1 _steel T = 60 mm - 59.99 mm(59.99 mm × 19 × 10^-6 /K) - (60 mm × 12 × 10^-6 /K) T = 0.01 mm(1139.81 × 10^-6 mm/K) - (720 × 10^-6 mm/K) T = 0.01 mm(1139.81 - 720) × 10^-6 mm/K T = (0.01)/(419.81 × 10^-6) K T ≈ 23.82 K Step 4: Calculate the final temperature (T_2). T_2 = T_1 + T T_2 = 303 K + 23.82 K T_2 = 326.82 K The common temperature at which the ball will just pass through the hole is 326.82 K. That's 2 down. 3 left today — send the next one.

An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11 x 10 -6/°C. Calculate: a) The area expansion coefficient of iron. b) The lengths of the sides at 51 °C. c) The new area at 51 °C.

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An iron plate is a 10 cm square at 12 °C. The linear expansion coefficient of iron is 11 x 10 -6/°C. Calculate: a) The area expansion coefficient of iron. b) The lengths of the sides at 51 °C. c) The new area at 51 °C.

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