The molar heat capacity of water at constant, pressure, $\mathrm{C}$, is $75 \mathrm{~J} \mathrm{~K}^{-1}…
- $1.2 \mathrm{~K}$
- $2.4 \mathrm{~K}$
- $4.8 \mathrm{~K}$
- $6.6 \mathrm{~K}$
Solution
$\begin{aligned}
& 18 \mathrm{~g} \text { of water }=1 \text { mole }=75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\
& 1 \mathrm{~g} \text { of water }=\frac{75}{18} \mathrm{JK}^{-1} \\
& \mathrm{Q}=\text { m.c. } \Delta t
\end{aligned}$
$\begin{aligned}
1000 & =100 \times \frac{75}{18} \times \Delta t \\
\Rightarrow \Delta t & =\frac{10 \times 18}{75}=2.4 \mathrm{~K}
\end{aligned}$
Asked in: NEET 2003