(A) The heat $Q$ required to change the temperature of a substance is given by $Q = mc\Delta T$.For copper: $m_c = 50 \, g = 0.05 \, kg$,$c_c = 420 \,

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(A) The heat $Q$ required to change the temperature of a substance is given by $Q = mc\Delta T$.For copper: $m_c = 50 \, g = 0.05 \, kg$,$c_c = 420 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$,$\Delta T_c = 10^\circ C$.$Q = 0.05 \times 420 \times 10 = 210 \, J$.This same heat is given to $10 \, g$ of water $(m_w = 0.01 \, kg)$.Using $Q = m_w c_w \Delta T_w$:$210 = 0.01 \times 4200 \times \Delta T_w$.$210 = 42 \times \Delta T_w$.$\Delta T_w = \frac{210}{42} = 5^\circ C$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Principle of Calorimetry and Water Equivalent

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The heat required to raise the temperature of $50 \, g$ of copper by $10^\circ C$ is given to $10 \, g$ of water. What is the rise in temperature of the water in $^\circ C$? (Specific heat of copper $= 420 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$,Specific heat of water $= 4200 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$)

A
$5$
B
$6$
C
$7$
D
$8$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

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$540 \ g$ of ice at $0 \ ^\circ C$ is mixed with $540 \ g$ of water at $80 \ ^\circ C$. The final temperature of the mixture is

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