Question

If $$3\text{g}$$ of steam at $$100{}^{\circ }\text{C}$$ is passed into a calorimeter containing a liquid then the temperature of the liquid rises from $$30{}^{\circ }\text{C}$$ to $$40{}^{\circ }\text{C}$$. Calculate the water equivalent of the calorimeter and its contents. L of vaporization of water=$$540\text{cal/g}$$. ($$180\text{g}$$)

Answer 1

Hint:Use the formula for heat exchanged by the substance with the surrounding in terms of change in temperature, specific heat and mass of the substance. Also use the formula for the heat exchanged in terms of latent heat of vaporization. Use the law of conservation of energy and calculate the required answer.

Formulae used:
The heat $$Q$$ exchanged by a substance is
$$Q=mC\mathrm{\Delta }T$$ …… (1)
Here, $$m$$ is mass of the substance, $$C$$ is specific heat of substance and $$\mathrm{\Delta }T$$ is change in temperature of the substance.
The amount of heat $$Q$$ absorbed or released by a substance is
$$Q=m{L}_V$$ …… (2)
Here, $$m$$ is the mass of the substance and $$L_V$$ is the latent heat of vaporization of the substance.

Complete step by step answer:
We have given that the mass of the steam is $$3\text{g}$$ and the temperature of the steam is $$100{}^{\circ }\text{C}$$.
$$m=3\text{g}$$
$$\Rightarrow T_s=100{}^{\circ }\text{C}$$
The initial and final temperature of the liquid in a calorimeter are $$30{}^{\circ }\text{C}$$ and $$40{}^{\circ }\text{C}$$ respectively.
$$\Rightarrow T_i=30{}^{\circ }\text{C}$$
$$\Rightarrow T_f=40{}^{\circ }\text{C}$$

The specific heat of water is $$1\text{cal}\cdot {\text{g}}^{-1}{\cdot }^{\circ }{\text{C}}^{\text{ - 1}}$$.
$$C=1\text{cal}\cdot {\text{g}}^{-1}{\cdot }^{\circ }{\text{C}}^{\text{ - 1}}$$
The total heat $$Q$$ lost by the steam is equal to the heat $$Q_l$$ lost by steam and heat $$Q_V$$ due to vaporization.
$$Q=Q_l+Q_V$$
Substitute $$mC\left(T_s-T_f\right)$$ for $$Q_l$$ and $$m{L}_V$$ for $$Q_V$$ in the above equation.
$$Q=mC\left(T_s-T_f\right)+m{L}_V$$

Substitute $$3\text{g}$$ for $$m$$, $$1\text{cal}\cdot {\text{g}}^{-1}{\cdot }^{\circ }{\text{C}}^{\text{ - 1}}$$ for $$C$$, $$100{}^{\circ }\text{C}$$ for $$T_s$$, $$40{}^{\circ }\text{C}$$ for $$T_f$$and $$540\text{cal/g}$$ for $$L_V$$ in the above equation.
$$Q_s=\left(3\text{g}\right)\left(1\text{cal}\cdot {\text{g}}^{-1}{\cdot }^{\circ }{\text{C}}^{\text{ - 1}}\right)\left(100{}^{\circ }\text{C}-40{}^{\circ }\text{C}\right)+\left(3\text{g}\right)\left(540\text{cal/g}\right)$$
$$\Rightarrow Q_s=180\text{cal}+1620\text{cal}$$
$$\Rightarrow Q_s=1800\text{cal}$$
Hence, the total heat lost by the steam is $$1800\text{cal}$$.

The heat $$Q_C$$ gained by the liquid and content in calorimeter is given by
$$Q_C=mC\left(T_f-T_i\right)$$
Let $$m_L$$ be the water equivalent of the liquid and content in a calorimeter.
Substitute $$1\text{cal}\cdot {\text{g}}^{-1}{\cdot }^{\circ }{\text{C}}^{\text{ - 1}}$$ for $$C$$, $$30{}^{\circ }\text{C}$$ for $$T_i$$ and $$40{}^{\circ }\text{C}$$ for $$T_f$$ in the above equation.
$$\Rightarrow Q_C=m_L\left(1\text{cal}\cdot {\text{g}}^{-1}{\cdot }^{\circ }{\text{C}}^{\text{ - 1}}\right)\left(40{}^{\circ }\text{C}-30{}^{\circ }\text{C}\right)$$
$$\Rightarrow Q_C=10m_L$$

According to the law of conservation of energy, the heat lost by the steam is equal to the heat gained by the liquid and contents in a calorimeter.
$$Q=Q_C$$
Substitute $$1800\text{cal}$$ for $$Q$$ and $$10m_L$$ for $$Q_C$$ in the above equation.
$$1800\text{cal}=10m_L$$
$$\therefore m_L=180\text{g}$$

Hence, the water equivalent of the liquid and content in the calorimeter is $$180\text{g}$$.

Note: The students should not forget to consider the heat lost due to vaporization of the steam. If this value is not considered then the final answer will not be correct. Also, there is no need to convert the unit of mass of the steam in the SI system of units as the final answer is in the CGS system of units.