Specific Heat Calculator 


               


Specific Heat Calculator


Specific heat


specific heat formula


Btu/hr
lb
°F

0.00 Btu/lb•°F




The specific heat calculator calculates the specific heat of a substance.

The specific heat is the amount of heat necessary to raise the temperature of 1lb of a substance 1°F.

Every substance has a different specific heat. The specific heat for water in the liquid state is 1 Btu/lb/°F. This means that 1lb of liquid water will be raised 1°F if 1 Btu is applied to it. The specific heat of aluminum is 0.224. This means that aluminum has a lower specific heat, meaning less heat is required to raise its temperature. Specific heat is important to know how easy or difficult it is to raise the heat of a substance. It is a measurement standard. The lower the specific heat of a substance, the easier it is to increase its temperature. The higher the specific heat, the more difficult it is to increase its temperature. Below is the chart of the specific heat of many other common substances.

To calculate the specific heat, the 3 parameters that need to be known are the quantity of heat applied to the substance, Q, the mass of the substance, m, and the change in the rise of temperature of the substance, ΔT.

The quantity of heat applied to the substance is Q. This is the amount of heat, in unit Btu, that is applied to the substance in order to raise its temperature.

The mass of the substance is its inherent mass in unit pounds (lbs).

The change of temperature, ΔT, is the rise temperature that the substances increases based on the Btu of heat applied to it. Therefore, if an object has a temperature rise of 30°F to 45°F, it has registered a 15°F rise. Therefore, ΔT= 15°F.

To use this calculator, a user enters in these 3 parameters and clicks the 'Calculate' button, and the resultant value will automatically be computed. The resultant specific heat value is in unit lb•°F.

Chart of the Specific Heat of Different Substances

Substance Specific Heat
(Btu/lb•°F)
Aluminum 0.224
Brick 0.22
Concrete 0.156
Copper 0.092
Ice 0.504
Iron 0.129
Marble 0.21
Steel 0.116
Water 1.00
Sea water 0.94
Air 0.24


Example

What is the specific heat of a substance that has 1000 Btu applied to it with a mass of 10lbs and a ΔT of 20°F?


c= Q/mΔT= 1000Btu/(10lbs * 20°F)= 0.5 Btu/lb°F