Question

The specific heat of aluminum is 0.0215 cal/g°c. if a 4.55 g sample of aluminum absorbs 2.55 cal of energy, by how much will the temperature of the sample change?

Answer 1

The formula we will use on this one is:

E = m C ΔT

where,

E = energy

m = mass of sample

C = specific heat

ΔT = change in temperature

 

Calculating for ΔT:

ΔT = E / m C

ΔT = 2.55 cal / (4.55 g * 0.0215 cal/g°C)

ΔT = 0.098 °C