Question

the specific heat of a substance which absorbs 7.5 × 10³j of heat when a sample with a mass of 1.0 × 10⁵g of the substance increases in temperature from 10°c to 90°c will be:
a 1.06 × 10³g°c/j
b 6.00 × 10¹⁰jg°c
c 6.00 jg/°c
d 9.4 × 10⁻⁴j/g°c

Explanation:

Step1: Recall the formula for specific heat

The formula for specific heat \(c=\frac{Q}{m\Delta T}\), where \(Q\) is the heat absorbed, \(m\) is the mass, and \(\Delta T\) is the change in temperature.

Step2: Calculate \(\Delta T\)

\(\Delta T=T_{final}-T_{initial}=90^{\circ}C - 10^{\circ}C=80^{\circ}C\)

Step3: Substitute the values into the formula

Given \(Q = 7.5\times10^{3}J\), \(m = 1.0\times10^{5}g\), \(\Delta T=80^{\circ}C\)
\(c=\frac{7.5\times 10^{3}J}{1.0\times 10^{5}g\times80^{\circ}C}\)
\(c=\frac{7.5\times 10^{3}}{8\times10^{6}}J/g^{\circ}C\)
\(c = 9.375\times10^{-4}J/g^{\circ}C\approx9.4\times 10^{-4}J/g^{\circ}C\)

Answer:

d. \(9.4\times 10^{-4}J/g^{\circ}C\)