# Heat energy required

1. Mar 11, 2009

### Damien H

Hi, wonder somone could help me, Im not physics expert. If a lube oil has a specific heat of 0.444Btu/Lb.F, how do I calculate the heat energy (kw) required to raise its temperature from 10C to 40C. Amount of liquid is 35L. Will I be able to calculate the time it takes to raise the temperature as well? someone told me its as easy as using the equation Q=CMΔT but another said I have to use Q=cpdT, Im confused!

2. Mar 11, 2009

### Marco_84

First i think you should specify wich phisycal quantities hide behind those letters :-),
in any case The specific heat capacity of a material is:

$$c={\partial C \over \partial m}$$

In absence of phase transition you have

$$c=E_ m={C \over m} = {C \over {\rho V}}$$

where:

C is the heat capacity of a body made of the material in question,
m is the mass of the body,
V is the volume of the body, and

$$\rho = \frac{m}{V}$$ is the density of the material.

I bet this is the relation in your's formula:

$$c_p=CM$$

where C is specific heat capacity at const pressure M the mass of the system and c_p is the body heat capacity at constant pressure.

http://en.wikipedia.org/wiki/Specific_heat_capacity

bye

marco

3. Mar 11, 2009

### Staff: Mentor

Use this one: