Question

$50g$ of copper is heated to increase its temperature by $10{}^{\circ }C$. If the same quantity of heat is given to $10g$ of water, then the rise in its temperature (in ${}^{\circ }C$ ) is:
(Specific heat of copper$=420Jk{g}^{-1}{K}^{-1}$,
Specific heat of water $=4186Jk{g}^{-1}{K}^{-1}$)

• A. 5

Answer 1

The correct option is A 5

Given:
Mass of copper, $m_c=50g=0.05kg$
Temperature change of copper coin, $\Delta T_c={10}^{\circ }C$
Specific heat of copper, $S_c=420Jk{g}^{-1}{K}^{-1}$
Mass of water, $m_w=10g=0.01kg$
Specific heat of water, $S_w=4186Jk{g}^{-1}{K}^{-1}$.

Let the temperature change of water be $\Delta T_w$
Since, heat given to the copper is equal to the heat gained by the water,
$m_c{S}_c\Delta T=m_w{S}_w\Delta T_w$
$0.05\times 420\times 10=0.01\times 4186\times \Delta T_w$
$\Delta T_w=5^{\circ }C$