Question

$250g$ of water at ${30}^{o}C$ is contained in a copper vessel of mass $50g$. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $5^{o}C$. Given : specific latent heat of fusion of ice $=336\times {10}^{3}Jk{g}^{-1}$, specific heat capacity of copper $=400Jk{g}^{-1}{K}^{-1}$, specific heat capacity of water $=4200Jk{g}^{-1}{K}^{-1}$

• A. $94.93g$
• B. $64.93g$
• C. $84.93g$
• D. $74.93g$

Answer 1

Answer: (D)

$74.93g$

Heat absorbed by ice is equal to heat absorbed to melt to water at $0^{o}C$ + Heat absorbed to rise temperature from $0^{o}C$ to $5^{o}C=m\times 336\times {10}^3+m\times 4200\times 5$

Heat released by water and vessel is $0.25\times 4200\times (30-5)+0.05\times 400\times (30-5)$

Equating above equations, $m=74.93\times {10}^{-3}kg=74.93g$