Question

50 g of copper is heated to increase its temperature by 10⁰C. If the same quantity of heat is given to 10 g of water, the rise in its temperature is (Specific heat of copper

$= 420Joule ⥂ - ⥂ kg^{- 1}{^\circ}C^{- 1}$)

• A. 5⁰C
• B. 6⁰C
• C. 7⁰C
• D. 8⁰C

Answer 1

Answer: (A)

5⁰C

Answer 2

Same amount of heat is supplied to copper and water so $m_{c}c_{c}\Delta T_{c} = m_{\omega}c_{\omega}\Delta T_{\omega}$

⇒ ∆T_ω = $\frac{m_{c}c_{c}\Delta T_{c}}{m_{\omega}c_{\omega}} = \frac{50 \times 10^{- 3} \times 420 \times 10}{10 \times 10^{- 3} \times 4200} = 5{^\circ}C$