$50 \mathrm{~g}$ 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…

Question

$50 \mathrm{~g}$ 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 $4200 \mathrm{Jkg}^{-10} \mathrm{C}^{-1}$ )

Aicharya · Accepted Answer

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 $50 \times 420 \times 10=10 \times 4200 \times \Delta t_2$ $\Delta t_2=5^{\circ} \mathrm{C}$

$50 \mathrm{~g}$ of copper is heated to increase its temperature by $10^{\circ} \mathrm{C}$. If the same…

Solution