What is the specific heat of lead if 345g of lead change from 45.0-72.0°C when
1215 J of heat are added

0.13J/g°C

Explanation:

Given parameters:

Mass  = 345g

Initial temperature  = 45°C

Final temperature  = 72°C

Quantity of heat  = 1215J

Unknown:

Specific heat of lead  = ?

Solution:

To solve this problem, we use the expression below:

       H  = mass x specific heat capacity x (Final - initial temperature)

Now insert the parameters and solve;

     1215  = 345 x Specific heat x (72  - 45)

     1215  = 9315 x specific heat

       Specific heat  = 0.13J/g°C

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.