Question

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

• A. $5^{\circ }C$
• B. $6^{\circ }C$
• C. $7^{\circ }C$
• D. $8^{\circ }C$

Answer 1

The correct option is A $5^{\circ }C$

Same amount of heat is supplied to copper and water so,
Heat absorbed by copper = heat absorbed by water
heat absorbed = mass $\times$ specific heat $\times$ temperature difference
$m_c{c}_c\Delta {\theta }_c=m_W{c}_W\Delta {\theta }_W$
where, $m_c$ = mass of copper
$m_W$ = mass of water
$c_c$ = specific heat of copper
$c_W$ = specific heat of water
$(\Delta \theta {)}_c$ = rise in temperature for copper
$(\Delta \theta {)}_W$ = rise in temperature for water
$\Rightarrow \Delta {\theta }_W=\frac{m_c{c}_c(\Delta \theta {)}_c}{m_W{c}_W}=\frac{50\times {10}^{-3}\times 420\times 10}{10\times {10}^{-3}\times 4200}=5^{\circ }C$