What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0C to 70.0°C?

Answer 1

q = mC∆T

q = heat = 2.5x10³ J

m = mass = 1x10⁴ g

C = specific heat = ?

∆T = change in temperature = 70.0 - 10.0 = 60ºC

Solving for C, we have...

C = q / (m)(∆T)

C = 2.5x10³ J / (1x10⁴ g)(60º)

C = 0.42 J/gº (to 2 significant figures)