Thermodynamics on a Midwinter Morning

Explanation

Application of Physical Laws

Since this phenomenon occurs within an insulated, sealed container with no heat exchange with the outside, the total energy of the system is conserved. In other words, the “heat gained by the air, $Q_{\text{gain}}$Qgain​,” and the “heat lost by the hot water (water), $Q_{\text{loss}}$Qloss​,” are equal.

$Q_{\text{gain}} = Q_{\text{loss}}$Qgain​=Qloss​

Deriving the Heat Changes for Each Substance

Let $T_3$T3​ be the overall temperature at thermal equilibrium.

1. Heat gained by the air $Q_{\text{gain}}$Qgain​ The air is heated from temperature $T_1$T1​ to $T_3$T3​ in a constant-volume process. $Q_{\text{gain}} = m_a c_v (T_3 - T_1)$Qgain​=ma​cv​(T3​−T1​)

2. Heat lost by the hot water $Q_{\text{loss}}$Qloss​ The hot water is cooled from temperature $T_2$T2​ to $T_3$T3​. $Q_{\text{loss}} = m_w c_w (T_2 - T_3)$Qloss​=mw​cw​(T2​−T3​)

Deriving $T_3$T3​ Using the Law of Conservation of Heat

Substitute each term into the equation for the conservation of heat.

$m_a c_v (T_3 - T_1) = m_w c_w (T_2 - T_3)$ma​cv​(T3​−T1​)=mw​cw​(T2​−T3​)

Rearrange this equation for $T_3$T3​.

$(m_a c_v + m_w c_w) T_3 = m_a c_v T_1 + m_w c_w T_2$(ma​cv​+mw​cw​)T3​=ma​cv​T1​+mw​cw​T2​

$T_3 = \frac{m_a c_v T_1 + m_w c_w T_2}{m_a c_v + m_w c_w}$T3​=ma​cv​+mw​cw​ma​cv​T1​+mw​cw​T2​​

Substituting Values and Calculating

Substitute the given constraint values.

• $m_a c_v = 35 \times 720 = 25200 \text{ J/K}$ma​cv​=35×720=25200 J/K
• $m_w c_w = 0.8 \times 4200 = 3360 \text{ J/K}$mw​cw​=0.8×4200=3360 J/K

The denominator is as follows. $m_a c_v + m_w c_w = 25200 + 3360 = 28560 \text{ J/K}$ma​cv​+mw​cw​=25200+3360=28560 J/K

The numerator is as follows. $m_a c_v T_1 + m_w c_w T_2 = 25200 \times 266 + 3360 \times 368$ma​cv​T1​+mw​cw​T2​=25200×266+3360×368 $= 6,703,200 + 1,236,480 = 7,939,680 \text{ J}$=6,703,200+1,236,480=7,939,680 J

Therefore, $T_3$T3​ is calculated as follows: $T_3 = \frac{7,939,680}{28,560} = 278 \text{ K}$T3​=28,5607,939,680​=278 K

Converted to Celsius, this temperature is $278 - 273 = 5 \text{ }^{\circ}\text{C}$278−273=5 ∘C. Since this exceeds the melting point of water, $T_0 = 273 \text{ K}$T0​=273 K, it satisfies the problem’s assumption that “water does not freeze and remains in a liquid state.” It divides evenly into an integer without any rounding or approximation.

The natural number to enter is 278.