Question

How much heat in joules is required to heat a 47 g sample of aluminum from 80 ∘F to 149 ∘F? (The specific heat capacity of aluminum is 0.903 J/(g⋅∘C).)

Answer 1

Given,

Mass of the sample, m=47 g= 0.047 kg

The initial temperature, T₁=80 °F=299.8 K

The final temperature of the sample, T₂=149 °F=338.2 K

Therefore the total raise in the temperature,

`\Delta T=T_2-T_1=338.2-299.8=38.4\text{ K}`

The specific heat capacity of the aluminum, c=0.903 J/(g · °C)= 903 J/(kg · K)

The heat required to raise the temperature of a sample by a temperature of ΔT is given by,

`Q=mc\Delta T`

On substituting the known values in the above equation,

`Q=0.047*903*38.4=1629.7\text{ J}`

Therefore the heat required to raise the temperature of the sample from 80 °F to 149 °F is 1629.7 J