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# 10.2: Specific Heat

Difficulty Level: At Grade Created by: CK-12
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This image is of the Beehive Geyser in Yellowstone National Park. Underground water is heated by the earth’s molten core and, when sufficient pressure is built up, the water shoots out of the ground in an amazing display.

### Specific Heat

When heat flows into an object, its thermal energy increases and so does its temperature. The amount of temperature increase depends on three things: 1) how much heat was added, 2) the size of the object, and 3) the material of which the object is made. When you add the same amount of heat to the same mass of different substances, the amount of temperature increase is different. Each substance has a specific heat, which is the amount of heat necessary to raise one mass unit of that substance by one temperature unit.

In the SI system, specific heat is measured in J/kg•K.  (Occasionally, you may also see specific heat expressed sometimes in J/g•K). The specific heat of aluminum is 903 J/kg•K. Therefore, it requires 903 J to raise 1.00 kg of aluminum by 1.00 K.

 Material Specific Heat (J/kg•K) Aluminum 903 Brass 376 Carbon 710 Copper 385 Glass 664 Ice 2060 Lead 130 Methanol 2450 Water Vapor 2020 Water (liquid) 4180 Zinc 388

The amount of heat gained or lost by an object when its temperature changes can be calculated by the formula

Q=mcΔt\begin{align*}Q = mc \Delta t\end{align*},

where Q\begin{align*}Q\end{align*} is the heat gained or lost, m\begin{align*}m\end{align*} is the mass of the object, c\begin{align*}c\end{align*} is its specific heat, and Δt\begin{align*}\Delta t\end{align*} is the change in temperature. You should note that the size of a Celsius degree and a Kelvin degree are exactly the same, and therefore Δt\begin{align*}\Delta t\end{align*} is the same whether measured in Celsius or Kelvin.

Example Problem:  A 0.500 kg block of zinc is heated from 295 K to 350. K. How much heat was absorbed by the zinc?

Solution:  Q=mcΔt=(0.500 kg)(388 J/kgK)(350. K295 K)=10,600 J\begin{align*}Q = mc \Delta t = (0.500 \ kg)(388 \ J/kg \cdot K)(350. \ K - 295 \ K) = 10,600 \ J\end{align*}

Example Problem:  845 J of heat are added to a 0.200 kg block of aluminum at a temperature of 312.00 K. How high will the temperature of the aluminum rise?

Solution: (t2t1)=Qmc=845 J(0.200 kg)(903 J/kgK)=4.68 K\begin{align*}(t_2-t_1)=\frac{Q}{mc}=\frac{845 \ J}{(0.200 \ kg)(903 \ J/kg \cdot K)}=4.68 \ K\end{align*}

t2=t1+4.68 K=312.00 K+4.68 K=316.68 K\begin{align*}t_2 = t_1 + 4.68 \ K = 312.00 \ K + 4.68 \ K = 316.68 \ K\end{align*}

#### Summary

• When heat flows into an object, its thermal energy increases and so does its temperature.
• The amount of temperature increase depends on three things:  1) how much heat was added, 2) the size of the object, and 3) the material of which the object is made.
• Each substance has a specific heat, which is the amount of heat necessary to raise one mass unit of that substance by one temperature unit.
• The amount of heat gained or lost by an object when its temperature changes can be calculated by the formula Q=mcΔt\begin{align*}Q = mc \Delta t\end{align*}.

#### Practice

Interactive step by step demonstration of how to calculate the specific heat of a material.

Practice problems in specific heat.

#### Review

1. How much heat is absorbed by 60.0 g of copper when it is heated from 20.0°C to 80.0°C?
2. A 40.0 kg block of lead is heated from -25°C to 200.°C.  How much heat is absorbed by the lead block?
3. The cooling system of an automobile motor contains 20.0 kg of water.  What is the Δt\begin{align*}\Delta t\end{align*} of the water if the engine operates until 836,000 J of heat have been added to the water?

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