Question

A 14.5 gram sample of metal was heated from 26 K to 71 K, a

temperature increase of 45.0 K, and absorbed 1432 J of energy as heat.
What is the specific heat of this metal?

Answer 1

Answer: -.433J/gC

Explanation: Q = mc x (Tfinal - T initial)

What you are given: 14.5 g sample of metal

temperature final 71K

Temp intial 26K

temperature change is 45 degrees K

Q is 1432 Joules

You are looking for c which is the specific heat, but that is Joules/g times degree Celcius

the temperature change is -228.16 degrees Celsius

so we know: 1432 = 14.5g x c x 45k

or 1432 = 14.5g x c x -228.16 C

C = 1432/ 14.5 x -228.16

C = -.433 J/gC