What is the specific heat of a substance if 2440 Joules are required to raise theÂ
  temperature of a 50g sample?
"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)"
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment