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

1 Answer
Jun 17, 2016

Answer:

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)#