Question

Find the result of mixing $10g$ of ice at $-{10}^{\circ }C$ with $10g$ of water at ${10}^{\circ }C$. Specific heat capacity of ice = $2.1J{g}^{-1}{K}^{-1}$, specific latent heat of ice = $336J{g}^{-1}$ and specific heat capacity of water = $4.2J{g}^{-1}{K}^{-1}$

• A. $6.25g$ ice will melt. Temperature will remain $0^{\circ }$
• B. $0.625g$ ice will melt. Temperature will increase to ${10}^{\circ }$
• C. $6.25g$ ice will melt. Temperature will increase to ${10}^{\circ }$
• D. $0.625g$ ice will melt. Temperature will remain $0^{\circ }$

Answer 1

Answer: (D)

$0.625g$ ice will melt. Temperature will remain $0^{\circ }$

heat realesed by water when it reaches to $0^{o}C$ is $Q=10\times 4.2\times 10=420J$

let the heat absorbed by ice , so the mass melt me m of ice

$420=m\times 336+10\times 2.1\times (0-(-10\left)\right)$

$m=0.625$, and system will remain at $0^{o}C$