Question

$5$g of water at ${30}^o$C and $5$g of ice at $-{20}^o$C are mixed in a calorimeter. Find the final temperature, water equivalent of calorimeter is negligible. Specific heat of ice is $0.5$ cal $g^{-1o}{C}^{-1}$ and latent heat of ice is $80$ cal per gram.

Answer 1

Given: 5 g of water at ${30}^{o}C$ and 5g of ice at −${20}^{o}C$ are mixed in a calorimeter, water equivalent of calorimeter is negligible. Specific heat of ice is $0.5cal/\left(g^{\circ }C\right)$ and latent heat of ice is 80 cal per gram.

To find the final temperature of the mixture

Solution:

We know,

Specific heat of water, $s_1=4.18J/\left(g^{\circ }C\right)=1cal/\left(g^{\circ }C\right)$

As per the given condition,

Mass of ice, $m_2=5g$

Mass of water, $m_1=5g$

Temperature of ice, $T_2=-{20}^{\circ }C$

Temperature of water, $T_1={30}^{\circ }C$

specific heat of ice, $s_2=0.5cal/g.°C$

latent heat of water $L=80cal{g}^{-1}.$

Let the final temperature be, $T$

Here, Heat lost by 5g water = 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)

$m_1{s}_1(T_1-T)=m_2{s}_2(0-T_2)+m_{2}L+m_2{s}_2(T-T_2)⟹5\times 1\times (30-T)=5\times 0.5(0-(-20\left)\right)+5\times 80+5\times 1\times (T-(0\left)\right)⟹150-5T=50+400+5T⟹10T=150-50-400⟹T=-{30}^{\circ }C$

So the final temperature of the mixture is ${30}^{\circ }C$