Question

Heat (in calorie) required to increase the temperature from \( 10^\circ C \) to \( 20^\circ C \) of 6 kg copper is the same as the heat (in calorie) required to increase the temperature from \( 20^\circ C \) to \( 100^\circ C \) of 3 kg lead. If the specific heat of copper is 0.09, then the specific heat of lead will be:

• A. 0.033
• B. 0.022
• C. 0.044
• D. 0.055

Hint:

To find the specific heat of a substance, use the heat equation \( Q = m \times c \times \Delta T \) and solve for \( c \) based on the equality of heat required for two different substances.

Answer 1

The Correct Option is B

The heat required for a change in temperature is given by the formula: \[ Q = m \times c \times \Delta T \] Where: - \( m \) is the mass of the substance - \( c \) is the specific heat capacity - \( \Delta T \) is the change in temperature For copper: \[ Q_{\text{copper}} = 6 \times 0.09 \times (20 - 10) = 6 \times 0.09 \times 10 = 5.4 \, \text{calories} \] For lead: \[ Q_{\text{lead}} = 3 \times c_{\text{lead}} \times (100 - 20) = 3 \times c_{\text{lead}} \times 80 \] Since the heat required for copper is equal to the heat required for lead: \[ 5.4 = 3 \times c_{\text{lead}} \times 80 \] Solving for \( c_{\text{lead}} \): \[ c_{\text{lead}} = \frac{5.4}{3 \times 80} = \frac{5.4}{240} = 0.0225 \, \text{cal/g}^\circ C \] Thus, the specific heat of lead is approximately \( 0.022 \, \text{cal/g}^\circ C \).