Question

A total of 54.0 calories of heat are absorbed as 58.3 g of lead is heated from 12.0° C to 42.0° C. What is the specific heat of lead? A) 0.029 cal gram °C B) 0.022 cal gram °C C) 0.017 cal gram °C D) 0.031 cal gram °C

Answer 1

The answer is choice D) 0.031 cal gram degrees celsuis

Answer 2

Answer: D) 0.031 cal / gram °C

Explanation:

Specific Heat can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 degree.

Q = m c ΔT

where m is the mass, c is the specific heat and ΔT is the difference in temperatures. Q = 54.0 cal

m = 58.3 g

ΔT = 42° C- 12°C = 30°C

c = Q / (mΔT)

c = 54.0 cal / (58.3 g×30°C) = 0.031 cal /gram °C