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$50$ g of copper is heated to increase its temperature by $10$ °C. If the same quantity of heat is given to $10$ g of water, the rise in its temperature is: (Specific heat of copper $= 420$ $J o u l e - k g^{- 1 \circ} C^{- 1}$ )

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$6^{\circ} C$

$7^{\circ} C$

$5^{\circ} C$

$8^{\circ} C$

answer is A.

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Detailed Solution

Same amount of heat is supplied to copper and water

$m_{c} c_{c} Δ T_{c} = m_{ω} c_{ω} Δ T_{ω}$

$Δ T_{ω} = \frac{m_{c} c_{c} Δ T_{c}}{m_{ω} c_{ω}} = \frac{50 \times 10^{- 3} \times 420 \times 10}{10 \times 10^{- 3} \times 4200} = 5 ° C$

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