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100 g of water is heated from $30^{0} C$ to $50^{0} C$ . Ignoring the, slight expansion of the water, the change in its internal enrgy is (Specific heat of water is 4184 J/Kg/K)

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4.2 KJ

8.4 KJ

84 KJ

2.1 KJ

answer is B.

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Detailed Solution

As work done = 0

W = PdV

dv = 0, w = 0

$Δ u = m c Δ T$

$= 100 \times 10^{- 3} \times 4184 \times (50 - 30)$

$= 100 \times 10^{- 3} \times 4184 \times 20$

$= 2 \times 4184 = 8.4 K J$

Change in internal energy = 8.4 KJ

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