Question

What is the specific heat of a substance if 2440 Joules are required to raise the 

  temperature of a 50g sample?

Answer 1

$c=\frac{Q}{m \Delta T}$

Q=2440 J

m=50 g=0.05 kg

$\Delta T=1 K$

$c=\frac{2440}{0.05(1)}=48800 J/(kg\ K)$