Question

Calculate the quantity of heat required to convert 1.5 kg of ice at ${100}^{\circ }C$ to water at ${15}^{\circ }C$. ($L_{ice}=3.34\times {10}^{5}J{kg}^{-1}$, $C_{water}=4180J{kg}^{-1}{}^{\circ }{C}^{-1}$)

• A. $5.85\times {10}^{5}J$
• B. $5.95\times {10}^{5}J$
• C. $3.95\times {10}^{5}J$
• D. $4.95\times {10}^{5}J$

Answer 1

Answer: (B)

$5.95\times {10}^{5}J$

We have , $L_{ice}=3.34\times {10}^{5}J/kg,m=1.5kg$

From the definition of latent heat , heat required to convert ice into water at constant temperature $0^{o}C$ ,

$Q_1=m{L}_{ice}$

or $Q_1=1.5\times 3.34\times {10}^5=5.01\times {10}^{5}J$

Now , heat required to heat up the water at $0^{o}C$ to${15}^{o}C$ ,

$Q_2=mc(15-0)=1.5\times 4180\times 15=0.94\times {10}^{5}J$ , where $c=4180J/kg{-}^{o}C$

Total heat required ,

$Q=Q_1+Q_2$

or $Q=5.01\times {10}^5+0.94\times {10}^5=5.95\times {10}^{5}J$