Question

A 4.0 g sample of a solid changes from 20.0 ºC to 36.0 ºC upon addition of 25.0 cal of heat. What is the specific heat of this solid?

Answer 1

Answer: c = 0.39 cal/g°C or

1.63 J/g°C

Explanation: To find the specific heat of the metal we will use the formula of heat which is Q= mc∆T.

We will derive for c, c = Q / m∆T

25.0 cal/ 4.0 g ( 36°C - 20°C)

= 0.39 cal / g°C

Or we can convert calories to joules.

0.39 cal x 4.184 J/ 1 cal

= 1.63 J /g°C