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, the rise in its temperature is:

(Given: ${\text{Specific heat of copper = 420 Joule kg}}^{-1}{}^{\circ }{C}^{-1}$, ${\text{Specific heat of water = 4200 Joule kg}}^{-1}{}^{\circ }{C}^{-1}$)

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

Answer 1

The correct option is A

$5^{\circ }C$

Amount of heat given to copper = Amount of heat given to water

$\Rightarrow$ $m_c{c}_c\Delta T_c=m_{\omega }{c}_{\omega }\Delta T_{\omega }$

$\Rightarrow \Delta T_{\omega }=\frac{m_c{C}_c\Delta T_c}{m_{\omega }{c}_{\omega }}=\frac{50\times {10}^{-3}\times 420\times 10}{10\times {10}^{-3}\times 4200}=5^{\circ }C$