A 2.0 imes 10^-2 ext kg ice at 0^ exto extC is dropped into a vacuum bottle originally holding 0.4 ext kg of water at 35^ exto extC. Assuming negligible heat loss or gain by the bottle, determine (a)

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Asked on May 3, 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 · 2026-05-03T08:25:15.111Z

You're on a roll — here are the solutions to questions 6, 7, 8, and 9. 6. A 2.0 × 10^-2 kg ice at 0^ is dropped into a vacuum bottle originally holding 0.4 kg of water at 35^. Assuming that any loss or gain of heat by the vacuum bottle is negligible, determine (a) the heat to melt the ice and (b) the final temperature after thermal equilibrium is attained. Given constants: Latent heat of fusion of ice, L_f = 3.34 × 10^5 J/kg Specific heat capacity of water, c_water = 4186 J/kg^ a) The heat to melt the ice Step 1: Use the formula for latent heat of fusion. The heat required to melt the ice is given by Q_melt = m_ice L_f. Q_melt = (2.0 × 10^-2 kg) × (3.34 × 10^5 J/kg) Step 2: Calculate the heat to melt the ice. Q_melt = 6.68 × 10^3 J The heat required to melt the ice is 6.68 × 10^3 J. b) The final temperature after thermal equilibrium is attained Step 1: Set up the heat exchange equation. Heat gained by ice (to melt and then warm up as water) = Heat lost by initial water. Let T_f be the final equilibrium temperature. Heat gained by ice: Q_gained = Q_melt + m_ice c_water (T_f - 0^) Heat lost by water: Q_lost = m_water c_water (35^ - T_f) Q_gained = Q_lost 6.68 × 10^3 J + (2.0 × 10^-2 kg)(4186 J/kg^)(T_f - 0^) = (0.4 kg)(4186 J/kg^)(35^ - T_f) Step 2: Simplify and solve for T_f. 6680 + 83.72 T_f = 1674.4 (35 - T_f) 6680 + 83.72 T_f = 58604 - 1674.4 T_f Combine terms with T_f: 83.72 T_f + 1674.4 T_f = 58604 - 6680 1758.12 T_f = 51924 T_f = (51924)/(1758.12) T_f ≈ 29.53^ The final temperature after thermal equilibrium is attained is approximately 29.53^. 7. What is the absolute temperature of boiling water and melting ice? Step 1: Recall the conversion formula from Celsius to Kelvin. The absolute temperature (in Kelvin) is found by adding 273.15 to the Celsius temperature: T_K = T_C + 273.15. Step 2: Calculate the absolute temperature for melting ice. Melting ice is at 0^. T_K = 0^ + 273.15 = 273.15 K Step 3: Calculate the absolute temperature for boiling water. Boiling water is at 100^. T_K = 100^ + 273.15 = 373.15 K The absolute temperature of melting ice is 273.15 K and boiling water is 373.15 K. 8. Determine the temperature whose Fahrenheit and Kelvin scales have the same reading. Step 1: Set up the equations for Fahrenheit and Kelvin in terms of Celsius. Let the temperature be x. So, T_F = x and T_K = x. The conversion from Celsius to Fahrenheit is T_F = (9)/(5)T_C + 32. The conversion from Celsius to Kelvin is T_K = T_C + 273.15, which means T_C = T_K - 273.15. Step 2: Substitute T_C into the Fahrenheit equation and set T_F = T_K = x. x = (9)/(5)(x - 273.15) + 32 Step 3: Solve for x. x = (9)/(5)x - (9)/(5)(273.15) + 32 x = 1.8x - 491.67 + 32 x = 1.8x - 459.67 Subtract 1.8x from both sides: x - 1.8x = -459.67 -0.8x = -459.67 x = (-459.67)/(-0.8) x = 574.5875 The temperature at which Fahrenheit and Kelvin scales have the same reading is approximately 574.59. 9. A scientist uses a scale for measuring temperature. In this scale water melts at 10^, and boils at 130^. The scientist measures the temperature at which sodium melts to be 127^. Express this temperature in Kelvin. Step 1: Establish the relationship between the scale and the Kelvin scale using the reference points. Water melts at 0^ or 273.15 K. Water boils at 100^ or 373.15 K. For the scale: T_, melt = 10^, T_, boil = 130^. For the Kelvin scale: T_K, melt = 273.15 K, T_K, boil = 373.15 K. The general formula for converting between two linear temperature scales is: T - T_lowerT_upper - T_lower = constant So, we can write: T_ - T_, meltT_, boil - T_, melt = T_K - T_K, meltT_K, boil - T_K, melt Step 2: Substitute the known values into the conversion formula. The temperature of sodium melting is T_ = 127^. (127 - 10)/(130 - 10) = (T_K - 273.15)/(373.15 - 273.15) (117)/(120) = (T_K - 273.15)/(100) Step 3: Solve for T_K. 0.975 = (T_K - 273.15)/(100) Multiply both sides by 100: 97.5 = T_K - 273.15 Add 273.15 to both sides: T_K = 97.5 + 273.15 T_K = 370.65 K The melting temperature of sodium in Kelvin is 370.65 K. What's next?

Accepted Answer · 2026-05-03T08:25:15.111Z

You're on a roll — here are the solutions to questions 6, 7, 8, and 9. 6. A 2.0 × 10^-2 kg ice at 0^ is dropped into a vacuum bottle originally holding 0.4 kg of water at 35^. Assuming that any loss or gain of heat by the vacuum bottle is negligible, determine (a) the heat to melt the ice and (b) the final temperature after thermal equilibrium is attained. Given constants: Latent heat of fusion of ice, L_f = 3.34 × 10^5 J/kg Specific heat capacity of water, c_water = 4186 J/kg^ a) The heat to melt the ice Step 1: Use the formula for latent heat of fusion. The heat required to melt the ice is given by Q_melt = m_ice L_f. Q_melt = (2.0 × 10^-2 kg) × (3.34 × 10^5 J/kg) Step 2: Calculate the heat to melt the ice. Q_melt = 6.68 × 10^3 J The heat required to melt the ice is 6.68 × 10^3 J. b) The final temperature after thermal equilibrium is attained Step 1: Set up the heat exchange equation. Heat gained by ice (to melt and then warm up as water) = Heat lost by initial water. Let T_f be the final equilibrium temperature. Heat gained by ice: Q_gained = Q_melt + m_ice c_water (T_f - 0^) Heat lost by water: Q_lost = m_water c_water (35^ - T_f) Q_gained = Q_lost 6.68 × 10^3 J + (2.0 × 10^-2 kg)(4186 J/kg^)(T_f - 0^) = (0.4 kg)(4186 J/kg^)(35^ - T_f) Step 2: Simplify and solve for T_f. 6680 + 83.72 T_f = 1674.4 (35 - T_f) 6680 + 83.72 T_f = 58604 - 1674.4 T_f Combine terms with T_f: 83.72 T_f + 1674.4 T_f = 58604 - 6680 1758.12 T_f = 51924 T_f = (51924)/(1758.12) T_f ≈ 29.53^ The final temperature after thermal equilibrium is attained is approximately 29.53^. 7. What is the absolute temperature of boiling water and melting ice? Step 1: Recall the conversion formula from Celsius to Kelvin. The absolute temperature (in Kelvin) is found by adding 273.15 to the Celsius temperature: T_K = T_C + 273.15. Step 2: Calculate the absolute temperature for melting ice. Melting ice is at 0^. T_K = 0^ + 273.15 = 273.15 K Step 3: Calculate the absolute temperature for boiling water. Boiling water is at 100^. T_K = 100^ + 273.15 = 373.15 K The absolute temperature of melting ice is 273.15 K and boiling water is 373.15 K. 8. Determine the temperature whose Fahrenheit and Kelvin scales have the same reading. Step 1: Set up the equations for Fahrenheit and Kelvin in terms of Celsius. Let the temperature be x. So, T_F = x and T_K = x. The conversion from Celsius to Fahrenheit is T_F = (9)/(5)T_C + 32. The conversion from Celsius to Kelvin is T_K = T_C + 273.15, which means T_C = T_K - 273.15. Step 2: Substitute T_C into the Fahrenheit equation and set T_F = T_K = x. x = (9)/(5)(x - 273.15) + 32 Step 3: Solve for x. x = (9)/(5)x - (9)/(5)(273.15) + 32 x = 1.8x - 491.67 + 32 x = 1.8x - 459.67 Subtract 1.8x from both sides: x - 1.8x = -459.67 -0.8x = -459.67 x = (-459.67)/(-0.8) x = 574.5875 The temperature at which Fahrenheit and Kelvin scales have the same reading is approximately 574.59. 9. A scientist uses a scale for measuring temperature. In this scale water melts at 10^, and boils at 130^. The scientist measures the temperature at which sodium melts to be 127^. Express this temperature in Kelvin. Step 1: Establish the relationship between the scale and the Kelvin scale using the reference points. Water melts at 0^ or 273.15 K. Water boils at 100^ or 373.15 K. For the scale: T_, melt = 10^, T_, boil = 130^. For the Kelvin scale: T_K, melt = 273.15 K, T_K, boil = 373.15 K. The general formula for converting between two linear temperature scales is: T - T_lowerT_upper - T_lower = constant So, we can write: T_ - T_, meltT_, boil - T_, melt = T_K - T_K, meltT_K, boil - T_K, melt Step 2: Substitute the known values into the conversion formula. The temperature of sodium melting is T_ = 127^. (127 - 10)/(130 - 10) = (T_K - 273.15)/(373.15 - 273.15) (117)/(120) = (T_K - 273.15)/(100) Step 3: Solve for T_K. 0.975 = (T_K - 273.15)/(100) Multiply both sides by 100: 97.5 = T_K - 273.15 Add 273.15 to both sides: T_K = 97.5 + 273.15 T_K = 370.65 K The melting temperature of sodium in Kelvin is 370.65 K. What's next?

A 2.0 imes 10^-2 ext kg ice at 0^ exto extC is dropped into a vacuum bottle originally holding 0.4 ext kg of water at 35^ exto extC. Assuming negligible heat loss or gain by the bottle, determine (a)

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A 2.0 imes 10^-2 ext kg ice at 0^ exto extC is dropped into a vacuum bottle originally holding 0.4 ext kg of water at 35^ exto extC. Assuming negligible heat loss or gain by the bottle, determine (a)

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