Question

A mixture of $250$g of water and $200$g of ice is kept in a calorimeter of water equivalent $50$g. If $200$g of steam at ${100}^o$C is passed through this mixture, calculate the final temperated and weight of contents of the calorimeter. Latent heat of steam is $540$ cal $g^{-1}$.

Answer 1

Given: A mixture of 250g of water and 200g of ice is kept in a calorimeter of water equivalent 50g. If 200g of steam at 100${}^o$C is passed through this mixture, Latent heat of steam is $540cal/g$.

To find the final temperature and weight of contents of the calorimeter.

Solution:

As per given criteria,

Mass of water, $m_w=250g$

Mass of ice, $m_i=200g$

Water equivalent, $w_e=50g$

Mass of steam, $m_s=200g$

Temperature of the steam, $T_s={100}^{\circ }C$

Latent heat of steam is $L=540cal/g$.

And we know,

Specific heat of water, $s=1cal/\left(g^{\circ }C\right)$

Latent heat of ice, $L_i=80cal/g$

According to the principle of calorimeter,

Heat lost = heat gained.

Let the final temperature of the mixture = 100ºC.

Heat gained by ice and iced water $Q_1$ is

$Q_1=m_i{L}_i+m_{i}s\Delta T=200\times 80+200\times 1\times 100=3600cal$.

$Q_2=(m_{w}s+w_e)\times (100-0)=(250\times 1+50)\times 100=30000cal$.

$Q=Q_1+Q_2=36000+30000=66000cal$.

If entire steam condensed the total heat given by it is

$Q_3=m_s\times L=200\times 540=108000cal$

Now the total heat available is 108000 cal, but only 66000 cal are required. Therefore all the steam will not get condensed.

Final temperature of the mixture is 100ºC.

Mass of steam condensed is $\frac{66000}{540}=122.2g$.

Mass of the contents is $250+200+122.2=572.2g$