Question

$1kg$ of water is supplied $630kJ$ of energy. Specific heat of water is $4200J/k{g}^{o}C$. Calculate the rise in temperature of $1kg$ water :

• A. ${40}^{o}C$
• B. ${60}^{o}C$
• C. ${80}^{o}C$
• D. ${150}^{o}C$

Answer 1

The correct option is D ${150}^{o}C$

The energy supplied ($Q$) to a substance of mass $m$ and specific heat $c$ is related to the rise in temperature ($\Delta t$) of the substance by the following relation: $Q=mc\Delta t$

Given: $m=1kg,Q=630kJ=630\times {10}^{3}J,c=4200J/k{g}^{o}C$ ,

Hence, by using above relation:

$\Rightarrow \Delta t=Q/mc=\frac{630\times {10}^3}{1\times 4200}={150}^{o}C$