Question

40 g of ice at 0${}^{\circ }C$ is used to bring down the temperature of a certain mass of water from 60${}^{0}C$ to 10${}^{0}C$. Find the mass of water used. Given, $S_w=4200Jk{g}^{-1}{}^{\circ }{C}^{-1}and{L}_f=336\times {10}^{3}jk{g}^{-1}$

• A. 72 g
• B. 38 g
• C. 44 g
• D. 40 g

Answer 1

The correct option is A

72 g

Mass of ice = 40 g = 0.04 kg

Let m kg of water is used.

Heat energy given by water = $m\times 4200\times (60-10)=210000mJ$

Heat energy taken by ice in melting $=0.04\times 336\times 1000=13440J$

Heat energy taken by molten ice to raise temperature $=0.04\times 4200\times 10=1680J$

Now, assuming no loss of energy

$210000m=15120(13440+1680)$

Hence, $m=0.072kg=72g$