Question

Find the result of mixing $10$ g of ice at $-{10}^{\circ }$C with $10$ g 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. $11.176$g
• B. $50.35g$
• C. $100.2g$
• D. $60.05g$

Answer 1

The correct option is A $11.176$g

Finally some ice will melt and become water and the whole mixture will reach a temperature of $0^{\circ }$C.

Let $x$ gm of ice will melt.

Heat energy gained by ice $=(x\times 2.1\times 10+x\times 336)$ J

Heat energy lost by water $=10\times 4.2\times 10=420$ J

since heat loss of ice equals heat gain of water,$x\times (21+336)=x\times 357=420$

Hence we get $x=1.176$ g

Hence final mixture is $11.176$ g of water $+8.824$ g of ice and the mixture is at temperature of $0^{\circ }$C