Question

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

Answer 1

0.13J/g°C

Step-by-step 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