The thermal cpacity of $11 k g$ of water is same as $120 k g$ of copper. If specific heat capacity of water is $4200 J k g^{- 1} K^{- 1}$ , find the specific heat capacity of copper.

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Solution

Step 1: Given data:
Specific heat capacity ( $s$ ) of water = $4200 J k g^{- 1} K^{- 1}$
Mass of water = $11 k g$
Mass of copper = $120 k g$
Step 2: Finding the specific heat capacity of copper:
We know that, $T h e r m a l c a p a c i t y = m \times s$
Where, $m$ is the mass.
$s$ is the specific heat capacity.
Hence, $m_{w} \times s_{w} = m_{C u} \times s_{C u}$
Where, $m_{w}$ is the mass of water
$m_{C u}$ is the mass of copper,
$s_{w}$ is the specific heat capacity of water
$s_{C u}$ is the specific heat capacity of copper.
Thus, $s_{C u} = \frac{m_{w} \times s_{w}}{m_{C u}}$
Substituting the known values,
$s_{C u} = \frac{11 \times 4200}{120} = 385 J k g^{- 1} K^{- 1}$
Hence, the specific heat capacity of copper $s_{C u} = 385 J k g^{- 1} K^{- 1}$ .

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