How much heat, in kJ, will be absorbed by a 25.0 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 25.0 °C to 76.0 °C?

Answer 1

Use q = mC∆T

q = heat = ?

m = mass = 25.0 g

C = specific heat = 0.930 J/gº

∆T = change in temperature = 76º - 25º = 51.0º

q = (25.0 g)(0.930 J/gº)(51.0º) = 1186 J = 1190 J (3 sig. figs.)