Question

Steam at 100°c is passed into 2kg of water contained in a calorimeter of water equivalent 0.02kg at 15°c till the temperature of the calorimeter and its contents rise to 90°c.The mass of steam condensed in kg is

Answer 1

Given :

Temperature of steam $T_1$ = 100°C

Mass of water M = 2 kg

Water equivalent of calorimeter m = 0.02 kg

Temperature of calorimeter $T_2$ = 15°C

Rise in temperature $T_3$ = 90°C

To find :

The mass of steam condensed = X kg

Solution :

As we know that

Latent heat of vaporization L = 540 cal kg⁻¹

Specific heat of water C = 1   $\frac{cal}{g\times \°C}$

By thermodynamic property,

$\textbf{\large Heat lost by steam =} \textbf{\large Heat gained by water and calorimeter}$

So,

$(X\times L) + (X\times C \times (T_1 - T_3)) = (M + m)\times C \times (T_3 - T_2)$

Putting all the values given above

$(X\times 540) + (X\times 1\times (100-90)) = (2+ 0.02) \times 1\times (90-15 )\\\\550X = 151.5$

So,

The mass of  steam condensed X = $\frac{ \textbf{\Large 151.5}}{ \textbf{\Large550}}$  kg

So, the mass of steam condensed = 0.275 kg