Question:medium

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:

Show 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.
Updated On: Jan 15, 2026
  • 0.033
  • 0.022
  • 0.044
  • 0.055
Show Solution

The Correct Option is B

Solution and Explanation

The formula to calculate the heat needed for a temperature change is: \[\nQ = m \times c \times \Delta T\n\] where: - \( m \) represents the mass. - \( c \) is the specific heat capacity. - \( \Delta T \) is the temperature change. For copper: \[\nQ_{\text{copper}} = 6 \times 0.09 \times (20 - 10) = 6 \times 0.09 \times 10 = 5.4 \, \text{calories}\n\] For lead: \[\nQ_{\text{lead}} = 3 \times c_{\text{lead}} \times (100 - 20) = 3 \times c_{\text{lead}} \times 80\n\] Given the heat required for copper equals the heat needed for lead: \[\n5.4 = 3 \times c_{\text{lead}} \times 80\n\] Solving for \( c_{\text{lead}} \): \[\nc_{\text{lead}} = \frac{5.4}{3 \times 80} = \frac{5.4}{240} = 0.0225 \, \text{cal/g}^\circ C\n\] Therefore, the specific heat of lead is approximately \( 0.022 \, \text{cal/g}^\circ C \).
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