Question

$400g$ of ice at 253 K is mixed with $0.05kg$ of steam at ${100}^{\circ }C$. Latent heat of vaporisation of steam is equal to 540 cal $g^{-1}$. Latent heat of fusion of ice is equal to 80 cal $g^{-1}$, Specific heat of ice is equal to 0.5 cal $g^{-1}$ ${}^{\circ }{C}^{-1}$. Find the resultant temperature of the mixture.

Answer 1

Given: 400g of ice at 253 K is mixed with 0.05kg of steam at ${100}^{\circ }C$. Latent heat of vaporisation of steam is equal to $540cal{g}^{-1}$. Latent heat of fusion of ice is equal to $80cal{g}^{-}1$, Specific heat of ice is equal to $0.5cal{g}^{-}{1}^{\circ }{C}^{-1}$.

To find the resultant temperature of the mixture.

Solution:

As per the given criteria,

mass of ice, $m_i=400g$

Temperature of ice, $T_i=253K=(253-273{)}^{\circ }C=-{20}^{\circ }C$

mass of the steam, $m_s=0.05kg=50g$

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

Latent heat of vaporisation of steam, $L_s=540cal{g}^{-1}$

Latent heat of fusion of ice, $L_f=80cal{g}^{-}1$

Specific heat of ice, $s=0.5cal{g}^{-}{1}^{\circ }{C}^{-1}$.

We know

Heat gained by ice=heat lost by steam

Heat energy needed to change the temperature of ice from –20°C to 0°C + Latent heat needed to change ice at 0°C into water at 0°C + heat absorbed by water (melted ice)=heat lost by steam

$m_{i}s(0-T_i)+m_i{L}_f+m_{i}s(T-T_i)=m_s{L}_s⟹400\times 0.5(0-(-20\left)\right)+400\times 80+400\times 0.5(T-(-20\left)\right)=50\times 540⟹4000+32000+20(T+20)=27000⟹20(T+20)=-9000⟹20T-400=-9000⟹20T=-9400⟹T=-{470}^{\circ }C$

So the final temperature of the mixture is $-{470}^{\circ }C$