Thermodynamics on a Midwinter Morning

Problem Statement

On a very cold midwinter morning, the room’s window was open, causing the indoor temperature to drop to below freezing. We close the window and trap a mass of air $m_a$ma​ inside a completely insulated, sealed container (which models this room). The initial temperature of the air at this point was $T_1$T1​.

To serve as a substitute for heating, a mass $m_w$mw​ of hot water at an initial temperature of $T_2$T2​ was gently placed inside the container, which was then resealed.

After a sufficient amount of time had elapsed, the air inside the room and the hot water reached thermal equilibrium, and the overall temperature became $T_3$T3​. In this process, we assume the following conditions:

• We consider only heat transfer from the hot water to the air; heat exchange with the outside and the heat capacity of the container can be neglected.
• Changes in the volume of the air can be neglected, so we treat the process as a constant-volume process. Let the specific heat at constant volume of the air be $c_v$cv​.
• Let the specific heat of water be $c_w$cw​.
• Let the melting point of water be $T_0$T0​. Assume that the water remains entirely in the liquid state without freezing and does not evaporate.

Derive the temperature $T_3$T3​ at thermal equilibrium using an algebraic expression, then substitute the given values of the constraints to find the value of $T_3$T3​.

Constraints

• Mass of air: $m_a = 35 \mathrm{ kg}$ma​=35kg
• Specific heat capacity of air at constant volume: $c_v = 720 \mathrm{ J/(kg\cdot K)}$cv​=720J/(kg⋅K)
• Initial temperature of air: $T_1 = 266 \mathrm{ K}$T1​=266K
• Mass of hot water: $m_w = 0.8 \mathrm{ kg}$mw​=0.8kg
• Initial temperature of hot water: $T_2 = 368 \mathrm{ K}$T2​=368K
• Specific heat of water: $c_w = 4200 \mathrm{ J/(kg\cdot K)}$cw​=4200J/(kg⋅K)
• Melting point of water: $T_0 = 273 \mathrm{ K}$T0​=273K

Input Format

Enter the value of the temperature $T_3$T3​ at thermal equilibrium as a natural number.