Calculate the heat gained by the calorimeter and water, given their masses, initial temperatures, specific heat capacities, and the final mixture temperature.

Question

Asked on May 1, 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 · 2026-05-01T13:05:21.874Z

Here's the solution to questions 13(c), 14(a), and 14(b). Question 13(c): Given: Mass of metal block, m_m = 150\, g = 0.15\, kg Initial temperature of metal block, T_m,i = 100\,^C Heat capacity of calorimeter, C_c = 40\, J/K Mass of water, m_w = 100\, g = 0.1\, kg Initial temperature of water, T_w,i = 25\,^C Final temperature of mixture, T_f = 34\,^C Specific heat capacity of water, c_w = 4200\, J/kg·K c) (i) Heat gained by calorimeter; Step 1: Calculate the change in temperature of the calorimeter. The calorimeter starts at the same temperature as the water. T_c = T_f - T_w,i = 34\,^C - 25\,^C = 9\,^C = 9\, K Step 2: Calculate the heat gained by the calorimeter using its heat capacity. Q_c = C_c T_c Q_c = (40\, J/K) (9\, K) Q_c = 360\, J The heat gained by the calorimeter is 360 J. c) (ii) Heat gained by water; Step 1: Calculate the change in temperature of the water. T_w = T_f - T_w,i = 34\,^C - 25\,^C = 9\,^C = 9\, K Step 2: Calculate the heat gained by the water using its mass and specific heat capacity. Q_w = m_w c_w T_w Q_w = (0.1\, kg) (4200\, J/kg·K) (9\, K) Q_w = 3780\, J The heat gained by water is 3780 J. c) (iii) Heat lost by the metal block; Step 1: Apply the principle of calorimetry: Heat lost by the metal block equals the total heat gained by the water and the calorimeter. Q_m = Q_w + Q_c Q_m = 3780\, J + 360\, J Q_m = 4140\, J The heat lost by the metal block is 4140 J. c) (iv) Specific heat capacity of the metal block Step 1: Calculate the change in temperature of the metal block. T_m = T_m,i - T_f = 100\,^C - 34\,^C = 66\,^C = 66\, K Step 2: Use the heat lost by the metal block to find its specific heat capacity, c_m. Q_m = m_m c_m T_m 4140\, J = (0.15\, kg) c_m (66\, K) 4140\, J = (9.9\, kg·K) c_m Step 3: Solve for c_m. c_m = 4140\, J9.9\, kg·K c_m = 418.1818... \, J/kg·K Rounding to one decimal place: c_m = 418.2\, J/kg·K The specific heat capacity of the metal block is 418.2 J/kg·K. Question 14: a) Distinguish between solid and liquid states of matter in terms of intermolecular forces. In solids, intermolecular forces are very strong, holding particles in fixed positions and giving solids a definite shape and volume. In liquids, intermolecular forces are weaker than in solids, allowing particles to move past each other, giving liquids a definite volume but no definite shape. b) In an experiment to estimate the diameter of an oil molecule, an oil drop of diameter 0.05\, cm spreads over a circular patch whose diameter is 20\, cm. b) (i) Determine the volume of the oil drop Step 1: Calculate the radius of the oil drop. r_drop = diameter of oil drop2 = 0.05\, cm2 = 0.025\, cm Step 2: Calculate the volume of the spherical oil drop. V_drop = (4)/(3) r_drop^3 V_drop = (4)/(3) × × (0.025\, cm)^3 V_drop = (4)/(3) × × 0.000015625\, cm^3 V_drop ≈ 6.54498 × 10^-5\, cm^3 Rounding to three significant figures: V_drop = 6.54 × 10^-5\, cm^3 The volume of the oil drop is 6.54 × 10^-5\, cm^3. b) (ii) Determine the area of the patch covered by the oil Step 1: Calculate the radius of the circular patch. r_patch = diameter of patch2 = 20\, cm2 = 10\, cm Step 2: Calculate the area of the circular patch. A_patch = r_patch^2 A_patch = × (10\, cm)^2 A_patch = × 100\, cm^2 A_patch ≈ 314.159\, cm^2 Rounding to three significant figures: A_patch = 314\, cm^2 The area of the patch covered by the oil is 314 cm^2. b) (iii) Determine the diameter of an oil molecule (implied by the context of the experiment) Step 1: Assume the oil spreads to form a monolayer, so the thickness of the patch is the diameter of one oil molecule, D_molecule. The volume of the oil drop is equal to the volume of the cylindrical patch. V_drop = A_patch × D_molecule Step 2: Solve for D_molecule. D_molecule = V_dropA_patch D_molecule = 6.54498 × 10^-5\, cm^3314.159\, cm^2 D_molecule ≈ 2.0833 × 10^-7\, cm Rounding to three significant figures: D_molecule = 2.08 × 10^-7\, cm The diameter of an oil molecule is 2.08 × 10^-7\, cm. 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-05-01T13:05:21.874Z

Here's the solution to questions 13(c), 14(a), and 14(b). Question 13(c): Given: Mass of metal block, m_m = 150\, g = 0.15\, kg Initial temperature of metal block, T_m,i = 100\,^C Heat capacity of calorimeter, C_c = 40\, J/K Mass of water, m_w = 100\, g = 0.1\, kg Initial temperature of water, T_w,i = 25\,^C Final temperature of mixture, T_f = 34\,^C Specific heat capacity of water, c_w = 4200\, J/kg·K c) (i) Heat gained by calorimeter; Step 1: Calculate the change in temperature of the calorimeter. The calorimeter starts at the same temperature as the water. T_c = T_f - T_w,i = 34\,^C - 25\,^C = 9\,^C = 9\, K Step 2: Calculate the heat gained by the calorimeter using its heat capacity. Q_c = C_c T_c Q_c = (40\, J/K) (9\, K) Q_c = 360\, J The heat gained by the calorimeter is 360 J. c) (ii) Heat gained by water; Step 1: Calculate the change in temperature of the water. T_w = T_f - T_w,i = 34\,^C - 25\,^C = 9\,^C = 9\, K Step 2: Calculate the heat gained by the water using its mass and specific heat capacity. Q_w = m_w c_w T_w Q_w = (0.1\, kg) (4200\, J/kg·K) (9\, K) Q_w = 3780\, J The heat gained by water is 3780 J. c) (iii) Heat lost by the metal block; Step 1: Apply the principle of calorimetry: Heat lost by the metal block equals the total heat gained by the water and the calorimeter. Q_m = Q_w + Q_c Q_m = 3780\, J + 360\, J Q_m = 4140\, J The heat lost by the metal block is 4140 J. c) (iv) Specific heat capacity of the metal block Step 1: Calculate the change in temperature of the metal block. T_m = T_m,i - T_f = 100\,^C - 34\,^C = 66\,^C = 66\, K Step 2: Use the heat lost by the metal block to find its specific heat capacity, c_m. Q_m = m_m c_m T_m 4140\, J = (0.15\, kg) c_m (66\, K) 4140\, J = (9.9\, kg·K) c_m Step 3: Solve for c_m. c_m = 4140\, J9.9\, kg·K c_m = 418.1818... \, J/kg·K Rounding to one decimal place: c_m = 418.2\, J/kg·K The specific heat capacity of the metal block is 418.2 J/kg·K. Question 14: a) Distinguish between solid and liquid states of matter in terms of intermolecular forces. In solids, intermolecular forces are very strong, holding particles in fixed positions and giving solids a definite shape and volume. In liquids, intermolecular forces are weaker than in solids, allowing particles to move past each other, giving liquids a definite volume but no definite shape. b) In an experiment to estimate the diameter of an oil molecule, an oil drop of diameter 0.05\, cm spreads over a circular patch whose diameter is 20\, cm. b) (i) Determine the volume of the oil drop Step 1: Calculate the radius of the oil drop. r_drop = diameter of oil drop2 = 0.05\, cm2 = 0.025\, cm Step 2: Calculate the volume of the spherical oil drop. V_drop = (4)/(3) r_drop^3 V_drop = (4)/(3) × × (0.025\, cm)^3 V_drop = (4)/(3) × × 0.000015625\, cm^3 V_drop ≈ 6.54498 × 10^-5\, cm^3 Rounding to three significant figures: V_drop = 6.54 × 10^-5\, cm^3 The volume of the oil drop is 6.54 × 10^-5\, cm^3. b) (ii) Determine the area of the patch covered by the oil Step 1: Calculate the radius of the circular patch. r_patch = diameter of patch2 = 20\, cm2 = 10\, cm Step 2: Calculate the area of the circular patch. A_patch = r_patch^2 A_patch = × (10\, cm)^2 A_patch = × 100\, cm^2 A_patch ≈ 314.159\, cm^2 Rounding to three significant figures: A_patch = 314\, cm^2 The area of the patch covered by the oil is 314 cm^2. b) (iii) Determine the diameter of an oil molecule (implied by the context of the experiment) Step 1: Assume the oil spreads to form a monolayer, so the thickness of the patch is the diameter of one oil molecule, D_molecule. The volume of the oil drop is equal to the volume of the cylindrical patch. V_drop = A_patch × D_molecule Step 2: Solve for D_molecule. D_molecule = V_dropA_patch D_molecule = 6.54498 × 10^-5\, cm^3314.159\, cm^2 D_molecule ≈ 2.0833 × 10^-7\, cm Rounding to three significant figures: D_molecule = 2.08 × 10^-7\, cm The diameter of an oil molecule is 2.08 × 10^-7\, cm. 3 done, 2 left today. You're making progress.

Calculate the heat gained by the calorimeter and water, given their masses, initial temperatures, specific heat capacities, and the final mixture temperature.

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Calculate the heat gained by the calorimeter and water, given their masses, initial temperatures, specific heat capacities, and the final mixture temperature.

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