Question

What is the specific heat of lead if 30.0 g of lead undergoes a 250.0°C change while absorbing 229.5 calories?

Answer 1

The equation here is `Q=c*m*DeltaT`

Explanation:

What we know:
`229.5=c*30.0*250.0->`
`c=229.5/(30.0*250.0)=0.0306cal//g*^oC`

Note:
It's more usual to use Joules for energy, where `1cal~~4.16J`
Also, the unit should be per `kg=1000g` and per Kelvin (`K`), but since it is a temperature change that doesn't make a difference:
`c=127.3J//kg*K=127.3J*kg^(-1)*K^(-1)`