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

Given \( Q = 4.537\space kJ\). Since \(1\space kJ=1000\space J\), then \(Q = 4.537\times1000 = 4537\space J\)

Step2: Use the heat - transfer formula \(Q = mc\Delta T\)

We know \(m = 200\space g\), \(c=0.790\space J/g^{\circ}C\), and \(Q = 4537\space J\). Rearranging the formula for \(\Delta T\) gives \(\Delta T=\frac{Q}{mc}\)
Substitute the values: \(\Delta T=\frac{4537}{200\times0.790}\)
First, calculate \(200\times0.790 = 158\)
Then, \(\Delta T=\frac{4537}{158}\approx28.7^{\circ}C\)

Answer:

\(28.7^{\circ}C\)