Question

If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0° C to 97.0° C, what mass is the sample of aluminum? (The specific heat of aluminum is 0.90 J/(g × ° C).)

Answer 1

Data:
Q (Amount of heat) = 832 J
m (mass) = ?
c (Specific heat) = 0.90 J/(g × ° C)
T (final) = 97 ºC
To (initial) = 20 ºC
ΔT = T - To → ΔT = 97 - 20 → ΔT = 77 ºC

Formula:
Q = m*c*ΔT

Solving:
Q = m*c*ΔT
832 = m*0.90*77
832 = 69.3m
69.3m = 832
$m = \frac{832}{69.3}$
$\boxed{\boxed{m \approx 12.00\:g}}\end{array}}\qquad\quad\checkmark$