Question

9. What is the resultant temperature when 100 g of steam at 100° C is passed through 500 g of
ice at -20° C. the specific heat of water is 0.5 cal g-1° C
-1
.

Answer 1

Answer on Question #42033 – Physics – Molecular Physics | Thermodynamics

Question.

What is the resultant temperature when $100\,\mathrm{g}$ of steam at $100{}^{\circ}\mathrm{C}$ is passed through $500\,\mathrm{g}$ of ice at $-20{}^{\circ}\mathrm{C}$. The specific heat of water is $0.5\,\mathrm{cal}\,\mathrm{g}^{-1}\mathrm{C}^{-1}$

Given:

$$\begin{array}{l} m_{1} = 100\,\mathrm{g} = 0.1\,\mathrm{kg} \text{ is a mass of steam} \\ m_{2} = 500\,\mathrm{g} = 0.5\,\mathrm{kg} \text{ is a mass of ice} \\ T_{1} = 100{}^{\circ}\mathrm{C} \text{ is an initial temperature of steam} \\ T_{2} = 0{}^{\circ}\mathrm{C} \text{ is an initial temperature of ice} \\ r = 2.26 \cdot 10^{6}\,\frac{\mathrm{J}}{\mathrm{kg}} \text{ is a heat of vaporization of steam} \\ \lambda = 0.334 \cdot 10^{6}\,\frac{\mathrm{J}}{\mathrm{kg}} \text{ is a heat of fusion of ice} \\ c = 1\,\frac{\mathrm{cal}}{\mathrm{g}{}^{\circ}\mathrm{C}} = 4200\,\frac{\mathrm{J}}{\mathrm{kg}{}^{\circ}\mathrm{C}} \text{ is a heat capacity of water} \end{array}$$

Find:

$$T = ? \text{ is a resultant temperature}$$

Solution.

The heat of condensation of steam and cooling the resulting water to a temperature $T$ is completely the heat of ice melting and heating the resulting water to a temperature $T$. In this case a mass of vapor $m_{1}$ and a mass of ice $m_{2}$ form a mass of water $(m_{1} + m_{2})$.

Write the heat balance equation:

$$Q_{1} = Q_{2}$$

$Q_{1}$ is a heat which steam gave

$Q_{2}$ is heat which ice got

$$Q_{1} = m_{1}r + m_{1}c(T_{1} - T)$$

$m_{1}r$ is a heat of condensing, steam converts to water

$m_{1}c(T - T_{1})$ is a heat of water cooling to the temperature $T$

$$Q_{2} = m_{2}\lambda + m_{2}c(T - T_{2})$$

$m_{2}\lambda$ is a heat of fusion, ice converts to water

$m_{2}c(T - T_{2})$ is a heat of water warming to the temperature $T$

So,

$$m _ {1} r + m _ {1} c \left(T _ {1} - T\right) = m _ {2} \lambda + m _ {2} c \left(T - T _ {2}\right)$$ $$m _ {1} r + m _ {1} c T _ {1} - m _ {1} c T = m _ {2} \lambda + m _ {2} c T - m _ {2} c T _ {2}$$ $$(m _ {1} + m _ {2}) c T = m _ {1} r - m _ {2} \lambda + m _ {1} c T _ {1} + m _ {2} c T _ {2}$$

Therefore,

$$T = \frac {m _ {1} r - m _ {2} \lambda + m _ {1} c T _ {1} + m _ {2} c T _ {2}}{(m _ {1} + m _ {2}) c} = \frac {m _ {1} r - m _ {2} \lambda}{(m _ {1} + m _ {2}) c} + \frac {m _ {1} T _ {1} + m _ {2} T _ {2}}{m _ {1} + m _ {2}}$$

Calculate:

$$\begin{array}{l} T = \frac {0 . 1 \cdot 2 . 2 6 \cdot 1 0 ^ {6} - 0 . 5 \cdot 0 . 3 3 4 \cdot 1 0 ^ {6}}{0 . 6 \cdot 4 2 0 0} + \frac {0 . 1 \cdot 1 0 0 + 0 . 5 \cdot 0}{0 . 6} = \frac {0 . 0 5 9 \cdot 1 0 ^ {6}}{2 . 5 2 \cdot 1 0 ^ {3}} + \frac {1 0 0}{6} = \\ = 2 3. 4 1 + 1 6. 6 7 = 4 0. 0 8 {}^ {\circ} \mathrm {C} \\ \end{array}$$

Answer.

$$T = \frac {m _ {1} r - m _ {2} \lambda}{(m _ {1} + m _ {2}) c} + \frac {m _ {1} T _ {1} + m _ {2} T _ {2}}{m _ {1} + m _ {2}} = 4 0. 0 8 {}^ {\circ} \mathrm {C}$$

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