Answer to Question #80390 in General Chemistry for Filly

Question

124.1g of an unknown substance absorbs 50.0kj of heat and increases its temperature by 36.3 degrees Celsius. What is the specific heat?

Answer 1

Solution:
Q = m Cp dT
where
Q = heat gained or lost = 50.0 kJ = 50,000 J
m = mass = 124.1 g
Cp = heat capacity
dT = change in temperature = 36.3 C
Cp = Q / (m dT)
Cp = (50,000 J) / (124.1 g x 36.3 C) = 11.1 J/gC
Answer: 11.1 J/gC