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13.5: Specific Heat

Difficulty Level: At Grade Created by: CK-12
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The student will:

  • Solve problems involving specific heat.


  • specific heat: The amount of heat required to raise the temperature of one gram of a substance by one Celsius degree.

Specific Heat

We know that when we heat an object, its temperature increases. But a given quantity of heat will not produce the same temperature change for every object. Recall that it requires 4.186 J of energy to raise the temperature of one gram of water by one degree Celsius. It can be determined experimentally that the same 4.186 J, if transferred to one gram of copper, will raise its temperature by 10.7 degrees Celsius.

For a given mass, experiments show that the temperature rise due to a fixed amount of energy transfer depends upon the material heated. The specific heat \begin{align*}c\end{align*} of a material is the amount of energy needed to raise the temperature of one kilogram of a particular substance by one-degree Celsius. The change in temperature \begin{align*}\Delta T\end{align*} of an object depends upon the amount of mass \begin{align*}m\end{align*} of the object being heated, the specific heat \begin{align*}c\end{align*} of the material the object is composed of, and the amount of energy \begin{align*}\Delta Q\end{align*} transferred to the object. It can be shown experimentally that:

\begin{align*}\Delta T=\frac{\Delta Q}{mc} \rightarrow \Delta Q=mc \Delta T\end{align*}

The SI units of specific heat are \begin{align*}\frac{J}{kg^\circ C}\end{align*}.

The Table below gives specific heat values for some common materials.

Substance Specific heat \begin{align*}\left(\frac{J}{kg^\circ C}\right)\end{align*} at standard atmospheric pressure and at 20°C
Liquid Water 4,186
Wood 1,700
Glass 840
Zinc 390
Copper 390
Silver 230


Check Your Understanding

One kilogram of the each of the substances in table 13.1.1 has 100 J of energy transferred to it at an initial temperature of 20°C. Which substance experiences the largest increase in temperature?

Answer: Since silver has the smallest specific heat, it has the largest temperature rise.

Illustrative Example 1

A United States penny, Figure below, is made almost entirely from zinc. The penny has a mass of 2.5 g and an initial temperature 20°C.

A United States penny is almost completely made from zinc

a. Find the temperature increase of the penny if 380 J is transferred to it.


Since the SI units of the specific heat contain kilograms, we must express 2.5 grams in kilograms:

\begin{align*}(2.5g) \frac{1.000kg}{1000g}=0.0025kg\end{align*}.

We should expect the temperature to increase to be greater than 1°C. 

\begin{align*}\Delta Q=cm \Delta T \rightarrow 380 J=\left(390 \frac{J}{Kg°C}\right)(0.0025kg) \Delta T \rightarrow\end{align*}

\begin{align*}\Delta T=\frac{380J}{\left(390 \frac{J}{Kg°C}\right)(0.0025kg)}=389.74 \rightarrow 390^\circ C\end{align*}

b. What is final temperature of the penny?


Since the penny was initially 20°C and the change in temperature was 390°C, the final temperature of the penny is 410°C. This is ten degrees away from its melting temperature.

Check Your Understanding

a. Equal amounts of zinc and silver, at the same initial temperature, absorbed the same amounts of heat. For every one degree increase in the temperature of the zinc, find the increase in the temperature of the silver.

Answer: The ratio of the specific heats of zinc to silver is \begin{align*}\frac{390 \frac{J}{kg^\circ C}}{230 \frac{J}{kg^\circ C}}=1.696 \rightarrow 1.70\end{align*}.

Therefore, the temperature of the silver will go up by \begin{align*}1.70^\circ C\end{align*} for every \begin{align*}1.00^\circ C\end{align*} increase in temperature of the zinc.

b. The temperature of a zinc coin increases from \begin{align*}20.0^\circ C\end{align*} to \begin{align*}25.0^\circ C\end{align*} after absorbing heat. What is the final temperature for a silver coin of the same mass at the same initial temperature which absorbs the same amount of heat?

Answer: \begin{align*}(5.0^\circ C)(1.696)=8.48 \rightarrow 8.5^\circ C\end{align*} increase in the temperature for the silver.  The final temperature is \begin{align*}28.5^\circ C\end{align*}.


If a system is insulated well for a period of time so that no appreciable energy is transferred beyond the system, we can think of such a system as being nearly “isolated.” This means for an isolated system composed of two objects with unlike temperatures, the hotter object will transfer energy to the colder object until the temperatures of both objects are identical.

Since energy must be conserved, we can find the final temperature of an isolated system. The energy lost by some of the objects must equal the energy gained by other objects. That is,

\begin{align*}-Q_{lost}=Q_{gained}, Q_{lost} <0\end{align*}

The energy lost by the hotter object is equal to the energy received by the colder object.

Illustrative Example 2

Let us imagine an isolated system consisting of a 30-g block of zinc heated to \begin{align*}300.0^\circ C\end{align*} which is placed into contact with a 30-g block of silver initially at a temperature of \begin{align*}20.0^\circ C\end{align*}.

What is the final temperature of the system, after equilibrium is reached?


The energy transferred by the zinc block is \begin{align*}-Q_{lost}=-c_{zinc}m_{zinc} \Delta T_{zinc}\end{align*} and the energy received by the silver block from is \begin{align*}Q_{gained}=c_{silver}m_{silver} \Delta T_{silver}\end{align*}.

Thus, \begin{align*}-c_{zinc}m_{zinc} \Delta T_{zinc} =c_{silver}m_{silver} \Delta T_{silver}\end{align*}. But \begin{align*}m_{zinc}=m_{silver}\end{align*}, so we can write:

\begin{align*}-c_{zinc} \Delta T_{zinc}=c_{silver} \Delta T_{silver}\end{align*}

\begin{align*}\Delta T_{zinc}=T_f-300.0^\circ C\end{align*} and \begin{align*}\Delta T_{silver}=T_f-20.0^\circ C\end{align*}

\begin{align*}-c_{zinc}(T_f-300.0^\circ C)=c_{silver}(T_f-20.0^\circ C) \rightarrow\end{align*}

\begin{align*}c_{zinc} T_f+c_{silver}T_f=(20.0^\circ C)c_{silver}+(300.0^\circ C)c_{zinc} \rightarrow\end{align*}

\begin{align*}T_f=\frac{(20.0^\circ C)c_{silver}+(300.0^\circ C)c_{zinc}}{c_{zinc}+c_{silver}}=\end{align*}

\begin{align*}\frac{(20.0^\circ C) \left(230 \frac{J}{kg^\circ C}\right)+(300^\circ C)\left(390 \frac{J}{kg^\circ C}\right)}{\left(390 \frac{J}{kg^\circ C}+230 \frac{J}{kg^\circ C}\right)} \rightarrow 196.13 \rightarrow 196^\circ C\end{align*}

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Date Created:
Mar 11, 2013
Last Modified:
Jul 28, 2016
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