Heat and Thermal ExpansionHard

Question

Density of water at $4^{\circ}C$ and $20^{\circ}C$ are $1000\text{ }kg/m^{3}$ 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 $= 4.2\text{ }J/kg$. and 1 atmospheric pressure $= 10^{5}\text{ }Pa$ )

Options

A.315826.2
B.234699.2
C.258700.8
D.268799.2

Solution

$Q = mS\Delta T = 4 \times 4200 \times 16\text{ }J = 268800\text{ }J$

$${W = P\Delta V }{\Delta V = \left( \frac{m}{\rho_{f}} - \frac{m}{\rho_{i}} \right) = 4\left\lbrack \frac{1}{998} - \frac{1}{1000} \right\rbrack }$$$P = 10^{5}\text{ }Pa$.

$${\therefore W = 10^{5} \times 4 \times \left\lbrack \frac{1}{998} - \frac{1}{1000} \right\rbrack = \frac{8 \times 10^{5}}{10^{3} \times 998} \approx 0.8\text{ }J }{\Delta U = Q - W = 268799.2\text{ }J}$$

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