Thermal Physics problem-solving question

On a hot day, the air temperature in a sealed, well-insulated bedroom is 24 °C. To cool the room down, a person takes a solid block of ice from the freezer, places it directly in front of an electric fan, and closes the door. The room measures 5.0 m × 4.0 m × 2.5 m. The ice block measures 0.25 m × 0.20 m × 0.10 m and is initially at 0 °C. Use the following data: density of air = 1.2 kg m^-3 specific heat capacity of air = 1000 J kg^-1 K^-1 density of ice = 920 kg m^-3 specific heat capacity of water = 4200 J kg^-1 K^-1 specific latent heat of fusion of ice = 3.3 × 10^5 J kg^-1 (a) Calculate the mass of air in the room. [2] (b) Calculate the mass of the ice block. [2] (c) The ice melts completely and the meltwater then warms up to the air temperature. Calculate the total energy absorbed by the ice and meltwater as it melts at 0 °C and then warms to 24 °C. [3] (d) Assume the room is perfectly sealed and insulated, and that all of the energy absorbed by the ice and meltwater is taken from the air. Calculate the drop in temperature of the air in the room. [2] (e) The fan's motor is rated at 60 W. Assume that all of the electrical energy supplied to the motor is eventually transferred to the air in the room as thermal energy. It takes 2.0 hours for the ice to melt completely. Calculate the temperature rise of the air caused by the motor running over this time. [2] (f) Using your answers and the idea of internal energy, evaluate the claim that placing ice in front of a fan is an effective way to cool a sealed room. In your answer, explain why running the fan alone (with no ice) would warm the room, and why a fan can still make a person feel cooler. [3]

(a) Volume of room V = 5.0 × 4.0 × 2.5 = 50 m^3. Mass of air m = ρV = 1.2 × 50 = 60 kg. (b) Volume of ice V = 0.25 × 0.20 × 0.10 = 5.0 × 10^-3 m^3. Mass of ice m = ρV = 920 × 5.0 × 10^-3 = 4.6 kg. (c) Energy to melt the ice (Q = ml): Q1 = 4.6 × 3.3 × 10^5 = 1.518 × 10^6 J. Energy to warm the meltwater 0 °C → 24 °C (Q = mcΔθ): Q2 = 4.6 × 4200 × 24 = 4.64 × 10^5 J. Total energy Q = Q1 + Q2 = 1.518 × 10^6 + 4.64 × 10^5 = 1.98 × 10^6 J ≈ 2.0 × 10^6 J. (d) Using Q = mcΔθ for the air, with Q = 1.98 × 10^6 J: Δθ = Q / (mc) = 1.98 × 10^6 / (60 × 1000) = 33 K (a fall of about 33 °C). (e) Energy transferred by the motor Q = Pt = 60 × (2.0 × 3600) = 60 × 7200 = 4.32 × 10^5 J. Temperature rise Δθ = Q / (mc) = 4.32 × 10^5 / (60 × 1000) = 7.2 K. (f) In this idealised model the ice removes far more energy than the motor adds (a 33 K fall against a 7.2 K rise), so the model predicts a net cooling of about 26 K. This is unrealistic: a real room is not perfectly sealed or insulated, so thermal energy flows in through the walls and windows and from people and other devices, and warm air leaks in — the actual temperature fall is much smaller. The air also cannot cool below 0 °C while ice is present. The motor does work and transfers energy to the air, which increases the internal energy (and hence temperature) of the air; with no ice, running the fan would slowly warm the room. A fan still makes a person feel cooler because the moving air increases the rate of evaporation of sweat from the skin and increases convection away from the body, lowering skin temperature without lowering the air...

(a) [2] - Volume of room = 50 m^3 (1) - Mass = ρV = 60 kg (1) (b) [2] - Volume of ice = 5.0 × 10^-3 m^3 (1) - Mass = ρV = 4.6 kg (1) (c) [3] - Energy to melt ice using Q = ml ≈ 1.5 × 10^6 J (1) - Energy to warm water using Q = mcΔθ ≈ 4.6 × 10^5 J (1) - Correct total ≈ 2.0 × 10^6 J (1) (d) [2] - Use of Δθ = Q / (mc) for the air (1) - Correct answer ≈ 33 K (1) (e) [2] - Use of Q = Pt with t = 7200 s (1) - Correct temperature rise ≈ 7.2 K (1) (f) [3] - Compares the cooling from the ice with the heating from the motor (net cooling in the model) (1) - States the real fall is much smaller, with a valid reason (room not sealed/insulated; energy gains from surroundings/people; air cannot fall below 0 °C) (1) - Fan alone warms the room because the motor transfers energy that raises the internal energy of the air; AND a fan feels cooler because of increased evaporation of sweat / convection from the skin (1)

Find each mass from m = ρV before any energy calculation. Melting needs Q = ml; changing temperature needs Q = mcΔθ. For the air, rearrange Q = mcΔθ to find Δθ. Energy from the motor is Q = Pt, with the time in seconds. In the evaluation, think about where energy in a real room comes from, and what the fan motor does to the air.