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Find the change in the internal energy of $2 k g$ of water as it is heated from $0 ° C$ to $4 ° C .$ The specific heat capacity of water is $4200 J / k g - K$ and its densities at $0 ° C$ and $4 ° C$ are $999.9 k g / m^{3}$ and $1000 k g^{'} m^{3}$ , respectively. Atmospheric pressure $= 10^{5} P a$ .

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$33600.02 J$

$25800.02 J$

$96300.02 J$

$11800.02 J$

answer is D.

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Detailed Solution

$V_{i} = \frac{m}{ρ_{t}} = \frac{2}{999.9} = 2.0002 \times 10^{- 3} m^{3}$

$V_{f} = \frac{m}{ρ_{f}} = \frac{2}{1000} = 2 \times 10^{- 3} m^{3}$

$Δ U = Q - W$

$= m s Δ θ - p_{d} Δ V$

$= (2) (4200) (4) - (10)^{5} (- 0.0002 \times 10^{- 3})$

$= 33600.02 J$

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