# A sample of an unknown substance has mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K, what is the specific heat of the substance?

Jun 17, 2016

Q = m c $\Delta$T
Rearrange to c = Q / (m $\Delta$T)

#### Explanation:

Q = 345.2J
m = 89.5 g
$\Delta$T = (305-285)= 20

So
c = 345.2 / (89.5 x 20)
c = 0.193 $J {g}^{-} 1 {K}^{-} 1$