Question

Calculate the quantity of heat required to convert $10kg$ of ice at $0^{o}C$ to water at ${50}^{o}C$ :

(Specific heat of water is $4200J/k{g}^{o}C$, Latent heat of fusion of ice is $336\times {10}^{3}J/kg$ )

• A. $4506kJ$
• B. $3360kJ$
• C. $5460J$
• D. $5460kJ$

Answer 1

The correct option is D $5460kJ$

We know that, to change the state of a matter, heat is required. Heat required to change a solid into liquid is given by ,

$Q=mL$ ,

where $L=$ latent heat of fusion ,

now given that , $m=10kg$ , and latent heat of fusion of ice is , $L=336\times {10}^{3}J/kg$ ,

hence, heat required to convert the 10kg of ice at $0^{o}C$ to water at $0^{o}C$ ,

$Q=(10\times 336\times {10}^3)=3360kJ$ ,

now,heat required to convert the 10kg of water at $0^{o}C$ to water at ${50}^{o}C$ ,

by , $Q^{\prime }=m\times c\times \Delta t$ ,

given $c=4200J/k{g}^{o}C$ ,

hence $Q^{\prime }=10\times 4200\times (50-0)=2100kJ$ ,

therefore total heat required to convert 10kg of ice at $0^{o}C$ to water at ${50}^{o}C$ ,

$Q^{\prime \prime }=Q+Q^{\prime }=3360+2100=5460kJ$