A total of 54.0 joules of heat are absorbed as 58.3 g of lead is heated from 12.0°C to 42.0°C. From these data, what is the specific heat of lead?

Answer 1

To find the specific heat of lead, we can use the equation:

q = mcΔT

Where q is the heat absorbed, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

Given data:
q = 54.0 J
m = 58.3 g
ΔT = 42.0°C - 12.0°C = 30.0°C

Plugging these values into the equation:

54.0 J = (58.3 g)(c)(30.0°C)

We can solve for c by dividing both sides of the equation by (58.3 g)(30.0°C):

c = 54.0 J / (58.3 g)(30.0°C)
c = 54.0 J / (1749 g°C)
c ≈ 0.0309 J/g°C

Therefore, the specific heat of lead is approximately 0.0309 J/g°C.