Question

question 2 (1 point)
a solid sample of granite, mass 200 g, is cooled such that it loses 4.537 kj of heat.
the specific heat capacity of granite is 0.790 j/g °c at satp. the reduction in
temperature is
38.7 °c
28.7 °c
48.7 °c
18.7 °c
58.7 °c

Explanation:

Step1: Convert heat loss to joules

The heat lost \( Q = 4.537\space kJ \). Since \( 1\space kJ = 1000\space J \), we convert:
\( Q = 4.537\times1000 = 4537\space J \)

Step2: Use the heat formula \( Q = mc\Delta T \) to solve for \( \Delta T \)

The formula for heat transfer is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. We rearrange for \( \Delta T \):
\( \Delta T = \frac{Q}{mc} \)

Substitute \( Q = 4537\space J \), \( m = 200\space g \), and \( c = 0.790\space J/g^\circ C \):
\( \Delta T = \frac{4537}{200\times0.790} \)

Calculate the denominator: \( 200\times0.790 = 158 \)

Then: \( \Delta T = \frac{4537}{158} \approx 28.7^\circ C \)

Answer:

28.7 °C (Option: 28.7 °C)