JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2026
- Hard
- MCQ
Density of water at \(4^{\circ} C\) and \(20^{\circ} C\) are \(1000 kg / m ^3\) and respectively. The increase in internal energy of 4 kg water when it is heated from \(4{ }^{\circ} C\) to \(20^{\circ} C\) is _________ J.
(Specific heat capacity of water \(=4.2 KJ / kg\). and 1 atmospheric pressure \(=10^5 Pa\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 268799.2
Step-by-step Solution
Detailed explanation
\( Q=mS\Delta T=4\times4200\times16~J=268800~J \)
\( W=P\Delta V \)
\( \Delta V=(\frac{m}{\rho_{f}}-\frac{m}{\rho_{i}})=4[\frac{1}{998}-\frac{1}{1000}] \)
\( P=10^{5}Pa. \)
.. \( W=10^{5}\times4\times[\frac{1}{998}-\frac{1}{1000}]=\frac{8\times10^{5}}{10^{3}\times998}\approx0.8~J \)
\( \Delta U=Q-W=268799.2~J \)
\( W=P\Delta V \)
\( \Delta V=(\frac{m}{\rho_{f}}-\frac{m}{\rho_{i}})=4[\frac{1}{998}-\frac{1}{1000}] \)
\( P=10^{5}Pa. \)
.. \( W=10^{5}\times4\times[\frac{1}{998}-\frac{1}{1000}]=\frac{8\times10^{5}}{10^{3}\times998}\approx0.8~J \)
\( \Delta U=Q-W=268799.2~J \)
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