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$50 \mathrm{~g}$ of copper is heated to increase its temperature by $10^{\circ} \mathrm{C}$. If the same quantity of heat is given to $10 \mathrm{~g}$ of water, the rise in temperature is (specific heat of $\mathrm{Cu}=420 \mathrm{Jkg}^{-10} \mathrm{C}^{-1}$ and specific heat of water is $4200 \mathrm{Jkg}^{-10} \mathrm{C}^{-1}$ )
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Verified Answer
The correct answer is:
$5^{\circ} \mathrm{C}$
Given,
Since, the same amount of heat supplied to copper and water
$Q_1=Q_2 \Rightarrow m_1 s_1 \Delta t_1=m_2 s_2 \Delta t_2$
Substituting the value, we get
$50 \times 420 \times 10=10 \times 4200 \times \Delta t_2$
$\Delta t_2=5^{\circ} \mathrm{C}$
Since, the same amount of heat supplied to copper and water
$Q_1=Q_2 \Rightarrow m_1 s_1 \Delta t_1=m_2 s_2 \Delta t_2$
Substituting the value, we get
$50 \times 420 \times 10=10 \times 4200 \times \Delta t_2$
$\Delta t_2=5^{\circ} \mathrm{C}$
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