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 is (specific heat of copper $=420J/kg{/}^{\circ }C$)

Answer 1

Given,
Mass of copper, $m_c=50g$
Rise in temperature, $\Delta T_c=10{}^{\circ }C$
Mass of water, $m_{\omega }=10g$

Same amount of heat is supplied to copper and water, so
$m_c{c}_c\Delta T_c$ = $m_{\omega }{c}_{\omega }\Delta T_{\omega }$
where, $m$, $c$ and $\Delta T$ is the mass, specific heat and change in temperature of copper and water respectively.

or, $(\Delta T{)}_{\omega }=\frac{m_c{c}_c\Delta T_c}{m_{\omega }{c}_{\omega }}$
$=\frac{50\times {10}^{-3}\times 420\times 10}{10\times {10}^{-3}\times 4200}$
$=5^{\circ }C$