Question

3 g of steam at 100$^\circ$C is passed into a calorimeter containing 20 g of a liquid. The temperature of the liquid rises from 30$^\circ$C to 40$^\circ$C. Calculate the water equivalent of the calorimeter and its contents. L of vaporization of water $= 540cal/g$.

Answer 1

We will use the principle of the calorimeter that amount of heat lost by the steam is equal to the amount of heat gained by the calorimeter. By calculating the amount of heat lost by steam to reach the temperature 40$^\circ$C and heat gained by a calorimeter to reach the same temperature and equating them, we can get the required answer.

Formula used:
The amount of heat energy released or absorbed at the boiling point of water is given in terms of latent heat of vaporization by the following expression:
$Q = mL$
The amount of heat energy released or absorbed by a substance is given by the following expression:
$Q = mc\Delta T$

Complete step by step answer:
The amount of steam given to us is
$m = 3g$
When steam is passed through the liquid then it will transfer its heat to the liquid by undergoing condensation. When change in state of steam takes place then latent heat of vaporization of water comes into action. Its value is given as
$L = 540cal/g$
The heat released by steam at 100$^\circ$C to change its state from steam to water is given as
$Q = mL = 3g \times 540cal/g = 1620cal$
Then after this, the temperature of water formed lowers from 100$^\circ$C to 40$^\circ$C. Therefore, for water we can write
$Q' = mC\Delta T = 3 \times 1 \times \left( {100 - 40} \right) = 3 \times 1 \times 60 = 180cal$
The specific heat capacity of water is $C = 1cal{\text{ }}{g^{ - 1}}{\text{ }}^\circ {C^{ - 1}}$.
The heat energy lost during this lowering in temperature is utilized in raising the temperature of the liquid from 30$^\circ$C to 40$^\circ$C.
The amount of heat absorbed by the liquid is given as
$Q'' = m'C\Delta T$
Here C is the specific heat capacity of the liquid and m’ is the water equivalent of the calorimeter and its contents. Inserting the known values, we get
$Q'' = m' \times 1 \times 10 = 10m'cal$
Now for a calorimeter, the amount of heat lost by steam is equal to the amount of heat gained by the calorimeter. Therefore, we can write that
$Q + Q' = Q'' \\\ \Rightarrow 1620 + 180 = 10m' \\\ \Rightarrow 10m' = 1800 \\\ \Rightarrow m' = 180g \\\$
This is the required value.

Note: The change in state of a material takes place at constant temperature. The heat supplied to the object at this temperature is utilized only for changing the state of the matter. The amount of heat required by that substance to change state at the given temperature is known as latent heat of that substance.