Specific Heat High School P. Science

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. Heat Energy

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 Capacity

The units most commonly used are joules per kilogram per degree Celsius (right side of table) Specific Heat Values

Specific Heat Values (J/g0C )

Specific Heat (cal/g0C) When given specific heat values, watch your units! Specific Heat (cal/g0C)

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) Heat Gained or Lost

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 A bomb calorimeter

A copper ornament has a mass of 0 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 J B 460 J C 540 J D 740 J Example Problem

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 final temperature of 27.4°C was used instead of the 7.4°C change in temperature. Explanation of Example

A 0. 0150 kg cylinder of zinc cooled from 100. 0°C to 20. 0°C 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°C C 559 J°/kg°C D 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. Explanation