Question

The temperature of a sample of iron with a mass of 10.0 g changed
from 50.4°C to 25.0°C with the release of 114.3 J of heat. What is the
specific heat of iron

Answer 1

Answer:

Q  =  mc(θ₂-θ₁)

47 calories =  10 g *c*(50.4 - 25)

47cal = 10*c* 25.4

47 /(10*25.4) = c

0.185 = c

Specific heat of iron = 0.185 cal/g°C

Explanation: