Chemistry 205

Question

A 28.07 g sample of a substance is initially at 22.6C. After absorbing 2605 J of heat, the temperature of the substance is 180.0C. What is the specific heat (SH) of the substance?

Answer 1

Use Q = mcΔT

 

Then c = Q / (mΔT)

 

m = 28.07 g

 

ΔT = 180.0°C - 22.6°C = 157.4°C

 

Q = 2605 J

 

c = Q / (mΔT) = (2605 J) / [(28.07 g)(157.4°C)] = 0.5896 J / (g °C)