Specific HeatHigh School P. Science

Heat Energy

Different substances have varying capacities for storing energy within their molecules.

Heat energy can cause molecules to move about faster, increasing their random kinetic energy.

An increase in this energy raises the temperature of the substance.

Heat Energy

Heat energy can also increase the vibrational or rotational energy of molecules, but this does not result in a temperature increase.

Specific Heat Capacity

Each substance has a unique specific heat capacity, meaning different substances have the ability to absorb only a certain amount of heat.

The specific heat capacity is generally defined as the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1°C.

It is a measure of how much heat energy a particular substance can hold.

Specific Heat Values

The units most commonly used are joules per kilogram per degree Celsius

(right side of table)

Specific Heat Values (J/g0C )

Specific Heat (cal/g0C)

When given specific heat values, watch your units!

Heat Gained or Lost

The amount of heat energy that a substance gains or loses, Q, depends on the mass (m), the specific heat, (c), and the change in the temperature (ΔT ) of the substance.

The formula for finding the heat energy is Q= mcΔT. (you will get this formula on the End of Course test)

A bomb calorimeter

Energy released is transferred to the water

The temperature change of the water is measured

Use known specific heat of water to determine energy transferred by the food sample to the water.

The amount of energy (heat) equals the energy content of the food

1 cal = 4, 184 J

Example Problem

A copper ornament has a mass of 0.0693 kg and changes from a temperature of

20.0°C to 27.4ºC. How much heat energy did it gain?

A 200 JB 460 JC 540 JD 740 J

Explanation of Example

The correct answer is choice A

Choice B is incorrect because the specific heat capacity of aluminum was used instead of the specific heat capacity of copper.

Choice C is incorrect. The initial temperature of 20.0°C was used instead of the 7.4°C change in temperature.

Finally, choice D is incorrect because the finaltemperature of 27.4°C was used instead of the

7.4°C change in temperature.

A 0.0150 kg cylinder of zinc cooled from 100.0°C to 20.0°C. The metal lost 466 J of heat energy. What is the specific heat capacity of the zinc?

A 311 J°/ kg°C B 388 J° /kg°CC 559 J°/kg°CD 1550 J°/kg°C

Explanation

choice B is the correct answer. Choice A is incorrect because the final

temperature of 100.0°C was used instead of ΔT = 80.0ºC.

Choice C is wrong because the incorrect formula, cP =QmΔT ,was used.Choice D is incorrect because the initial temperature of 20.0°C was used instead of ΔT = 80.0ºC.