Heat energy is given to $150 g$ of water, such that its temperature rises by $20 K$ . When the same amount of heat energy is given to a liquid X of mass $100 g$ its temperature rises by $80 K$ . Calculate
(a) heat energy given to water
(b) the specific heat capacity of liquid X. [Take up heat capacity of water $= 4.2 J g^{- 1} K^{- 1}$ ]

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Solution

Step 1: Given data
Water: mass $(m) = 150 g$
specific heat capacity $(c) = 4.2 J g^{- 1} K^{- 1}$
Temperature change $(θ_{R}) = 20 K$
Liquid X: mass $(m) = 100 g$
specific heat capacity (c)=?
temperature change $(θ_{F}) = 80 K$
Step 2: (a) heat energy given to water
Heat energy is given to water $= m c θ_{R}$
$= 150 g \times 4.2 J g^{- 1} K^{- 1} \times 20 K = 12600 J$
Step 3: (b)calculating the sp. heat capacity of liquid X
Heat energy is given to liquid X $= 12600 J$
$m c θ_{F} = 12600 J \Rightarrow 100 g \times c \times 80 K = 12600 J \Rightarrow c = \frac{12600}{8000} = 1.575 J g^{- 1} K^{- 1}$
Thus the heat energy given to the water is $12600 J$ and the specific heat capacity of liquid X is $1.575 J g^{- 1} K^{- 1}$ .

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