Question

A copper vessel of mass 100g contains 150g of water at ${50}^{o}C$. How much ice is needed to cool it to $5^{o}C$?
Given: Specific heat capacity of copper $=0.4J{g}^{-1}$ ${}^o{C}^{-1}$
Specific heat capacity of water $=4.2J{g}^{-1}$ ${}^o{C}^{-1}$
Specific latent heat of fusion of ice $=336J{g}^{-1}$

Answer 1

Heat given by water to reach $5^{o}C+$ Heat given by copper vessel to reach $5^{o}C=$ Heat taken by ice to melt at $0^{o}C+$ Heat taken by melted ice to reach $5^{o}C$

$m_w{c}_w\Delta t+m_c{c}_c\Delta t=m_{i}L+m_i{c}_w\delta t$

where $\Delta t=$ change in temperature of water and copper vessel

$m_w$ = mass of water = $150g$

$m_c$ = mass of copper vessel = $100g$

$m_i$ = mass of ice

$c_c$ = specific heat capacity of copper

$c_w$ = specific heat capacity of water

$L$ = specific latent heat of fusion of ice

$\delta t$ = change in temperature of ice

Therefore, $[150\times 4.2\times (50-5\left)\right]+[100\times 0.4\times (50-5\left)\right]=(m\times 336)+[m\times 4.2\times (5-0\left)\right]$

$(150\times 4.2\times 45)+(100\times 0.4\times 45)=(m\times 336)+(m\times 4.2\times m)$

$28350+1800=336m+21m$

$357m=30150$

$m=\frac{30150}{357}=84.45g$ of ice

Hence, $84.45g$ of ice is needed to cool it to $5°C$