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Heat capacity of water is 75 JK^-1 mole^-1. 3KJ of heat supplied to 180 grams of water rises the temperature by

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answer is D.

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Detailed Solution

q = 3000 J
$\begin{array}{l} C = 75 \times \frac{180}{18} {JK}^{- 1} \\ CΔT = q \\ ΔT = \frac{3000 J}{750 {JK}^{- 1}} = 4 K \end{array}$

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