Question

$50g$of copper is heated to increase its temperature by${10}^{0}C.$If the same quantity of heat is given to$10g$of water, the rise in its temperature is.$(specificheatofcopper=420J{K}^{-1}oC-1)$

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

Answer 1

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

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 top copper and water; so

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

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