Question

In a calorimeter of water equivalent 20 g ,water of mass 1.1 kg at 288 k temperature .if steam at teperature 373 k is passed through it and temperature of water increase by 6.5 degree c then the massss of steam condensed is ??????????????????????????????????

Answer 1

Answer on Question #37027, Thermodynamics

In a calorimeter of water equivalent 20 g, water of mass 1.1 kg at 288 k temperature. If steam at temperature 373 k is passed through it and temperature of water increase by 6.5 degree c then the mass of steam condensed is

Solution

Let $x$ be the mass of steam condensed. The heat obtained calorimeter with water is equal the heat from condensing $x$ kg of steam and the heat from cooling $x$ kg of water to the final temperature of calorimeter with water.

$$x \cdot L + x \cdot c \cdot (T_3 - T_2) = (m + m_c) \cdot c \cdot (T_2 - T_1),$$

where $L = 540\frac{cal}{g}$ - latent heat of steam, $c = 1\frac{cal}{g \cdot degree}$ - specific heat of water, $T_1 = 288\mathrm{k}$ - initial temperature of calorimeter with water, $T_2 = 288\mathrm{k} + 6.5\mathrm{K} = 294.5\mathrm{K}$ - final temperature of calorimeter with water, $m = 1.1\mathrm{kg}$ - mass of water, $m_c$ - mass of calorimeter, $T_3 = 373\mathrm{k}$ - temperature of steam.

The mass of steam condensed

$$x = \frac{(m + m_c) \cdot c \cdot (T_2 - T_1)}{L + c \cdot (T_3 - T_2)}.$$ $$x = \frac{(1.1 \cdot 10^3 g + 20 \mathrm{g}) \cdot 1 \frac{cal}{g \cdot degree} \cdot 6.5 \text{ degree}}{540 \frac{cal}{g} + 1 \frac{cal}{g \cdot degree} \cdot (373 \mathrm{k} - 294.5 \mathrm{K})} = 12g = 0.12 kg.$$

Answer: 0.12 kg.