Specific Heat Calculator is used to calculate the amount of energy required to raise temperature of unit mass (1 gram) of a pure substance by 1 degree. Specific heat tells us the measure of how thermally insensitive a substance is to the addition of energy.
Q = c m ΔT
Example 1: Find the specific heat of 350 g of a material when 34,700 Joules of heat are applied, and the temperature rises from 22ºC to 173ºC.
For the given problem:
m = mass = 350g
Q = energy = 34700 joules
ΔT = change in temperature = 173-22 = 151? C
We need to find the c specific heat. Now Since Q = cmΔT, therefore c = Q/mΔT
c = 34700/(350 x 151) = 34700/52850 = 0.65657521286 J/(g x ?C)
Example 2: Find the final temperature of 400 g of a material when 24,000 Joules of heat are applied, and the initial temperature was 20 ?C. The specific heat is 7.5 J/(g?C)
Q = energy = 24000 joules
ΔT = change in temperature = x-20 =(x–20)? C
c = Specific heat =7.5 J/(g?C)
We need to find the final temperature. Now Since Q = cmΔT, therefore ΔT = Q/mc
ΔT =24000/(400x7.5)=80 ?C.
Final temperature x=80+20 = 100?C.