Question

What would be the final temperature of the mixture when $5$g of ice at $-{10}^o$C are mixed with $20$g of water at ${30}^o$C. Specific heat of ice is $0.5$ and latent heat of water $80$cal $g^{-1}$.

Answer 1

Given: 5 g of ice at $-{10}^{o}C$ are mixed with 20g of water at ${30}^{o}C$. Specific heat of ice is 0.5 and latent heat of water $80cal{g}^{-1}.$

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, $m_1=20g$

Temperature of ice, $T_2=-{10}^{\circ }C$

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 20g water = Heat energy needed to change the temperature of ice from $–10°C$ to $0°C$ + Latent heat needed to change ice at 0°C into water at 0°C + heat absorbed by water (melted ice)

$m_1{s}_1(T_1-T)=m_2{s}_2(0-T_2)+m_{2}L+m_2{s}_2(T-T_2)⟹20\times 1\times (30-T)=5\times 0.5(0-(-10\left)\right)+5\times 80+5\times 1\times (T-(0\left)\right)⟹600-20T=25+400+5T⟹25T=600-400-25⟹T=7^{\circ }C$

is the final temperature.