Question

A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o
c. What is the specific heat of sc in j/g· o c?

Answer 1

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = $61.1^{0}C$

Heat absorbed by Scandium = $1.50*10^{3}J$

The equation showing the relationship between heat, mass, specific heat and temperature change:

$Q = m C (deltaT)$

Where Q is heat = $1.50*10^{3}J$

m is mass = 50.0 g

ΔT = $61.1^{0}C$

On plugging in the values and solving for C(specific heat) we get,

$1.50*10^{3}J$=50.0g(C)($61.1^{0}C$)

C = 0.491$\frac{J}{g^{0}C }$

Specific heat of the metal = 0.491$\frac{J}{g^{0}C }$