Data:
Q (Amount of heat) = ?(in Joule)
m (mass) = 35 g
c (Specific heat of iron) = 0.45 J/(g × ° C)
T (final) = 79 ºC
To (initial) = 14 ºC
ΔT = T - To → ΔT = 79 - 14 → ΔT = 65 ºC
Formula:
Q = mcΔT
Solving:
Q = mcΔT
Q = 350.4565
\(\boxed{\boxed{Q = 1023.75\:Joules}}\end{array}}\qquad\quad\checkmark\)
