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,
[tex]\Delta T=T_2-T_1=338.2-299.8=38.4\text{ K}[/tex]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,
[tex]Q=mc\Delta T[/tex]On substituting the known values in the above equation,
[tex]Q=0.047\times903\times38.4=1629.7\text{ J}[/tex]Therefore the heat required to raise the temperature of the sample from 80 °F to 149 °F is 1629.7 J
