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# Temperature and Heat Study Guide (page 3)

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Updated on Sep 27, 2011

#### Example

An aluminum piece of 400 g is placed in a container that holds 100 g of ice at –10°C . The water warms up to 5°C. In the process, the aluminum piece gets colder and reaches a temperature of 45°C. What was the initial temperature of the aluminum piece? The experiment takes place in a calorimeter.

#### Solution

First, convert all quantities to SI units. Next, set the equations and solve for the initial temperature.

According to the problem, the exchange happens in a calorimeter. Therefore, the only heat exchange happens between the aluminum and the ice.

mal = 400 g = 400 g ·1 kg/1,000 g = 0.4 kg
mice = 100 g = 100 g ·1 kg/l,000 g = 0.1 kg
tice initial = – 10° C
twater final = 5°C
tAl final = 45° C
tal initial = ?

Because the system is thermally isolated, the heat released by the aluminum is absorbed by the ice and then the water.

Qice = –Qaluminum

The process of heating of water goes through three different steps: First, the ice absorbs heat and the temperature increases from –10 to 0° C; second, the ice melts and there is no change in temperature; and last, the water warms up from 0 to 5° C. So we will have three terms referring to the ice to water heat absorption.

Qice = Q– 10 to 0° C + Latent heat of melting + Q0 to 5° C
Qice = mice · Cice ·(0°Ctice initial) + m · L + mice · Cwater · (twater final – 0° C)
(L = 3.33 × 105 J/kg)
Qice = mice · Cice · (0 – ( –10)) + mice · L + mice · Cwater · (5 – 0)
Qice = 3.74 . 104J
Qaluminum = ma1 · cal · (tal finaltal intial)
Qaluminum = 0.4 kg . 900 J/kg · °C· (45 – tAl initial)
3.74 · 104J = – 0.4 · 900/J°C · (45 – tAl initial)
3.74 · 104 J/(0.4 · 900J/°C) = – (45 – tAl initial)
3.74 · 1040 C/(0.4 · 900) = –(45 – tAl initial)
104°C = (45 – tAl initial)
tAl initial = (45 +104)°C = 149°
tAl initial = 149°C

## Thermal Expansion

We have now discussed the effect of heat exchange on a solid, and we learned that in the process, the particles inside a solid will move apart from each other as they acquire kinetic energy (due to an increase of average speed). The effect is easily observed if you make a careful measurement of the length of a rod that is subjected to increased temperature. This process of expansion is dependent not only on the amount of energy exchanged but also on the nature of the material.

In order to have a quantitative explanation of how substances expand, a few characteristics should be defined and measured: ΔL = LL0 = change in length; α = coefficient of thermal expansion (1/°C); L0 = initial length; ΔT = change in temperature.

The expansion along one direction is called linear thermal expansion and can be related to these coefficients and parameters through the following expression:

ΔL = α · L0 · ΔT

But the structure of a solid is three dimensional so we cannot assume that the process of thermal expansion happens linearly. All x, y, and z directions will be affected by the expansion. A simple way to correlate 3–D expansion with linear expansion is to imagine a cube of initial side L0 that is heated up to a temperature To + ΔT. All the sides of the cube will expand, and the change in volume be found by subtracting the initial and final volumes. The volume thermal expansion then is characterized by the following equation:

ΔV = β · V0 Δ T

where ΔV = change in volume; β = coefficient of volume thermal expansion; V0 = initial volume; ΔT = change in temperature. If we consider that:

ΔV = VV0 = L3L03

and we write an expression for L as a function of L0 as given by

ΔL = LL0 = α · L0 · Δ T

then we can show that by neglecting higher order terms in α (measurement shows this coefficient being very small; therefore, the higher orders will be even smaller. They can be neglected):

β = 3 · α

And:

ΔV = 3 · α · Vo · ΔT

#### Example

A copper cube of size 25 cm has a coefficient of linear expansion of 17 · 10 – 6° C–1 Find the change in length and volume if the temperature changes 12 °C from morning to noon.

#### Solution

First, convert the units to SI. Next, build our equations to determine the unknown.

α = 17 · 10 – 6° C–1
ΔT = 12°C
L0 = 25 cm = 0.25 m
ΔL = ?
ΔV = ?

We can determine first the change in length:

ΔL = α · L0 · Δ T =
17 · 10 – 6° C–1 · 0.25 m · 12°C = 5.1 · 10–5m
ΔL = 5.1 · 10–5m
ΔV = 3 · α · V0 · Δ T
V0 = L03
ΔV = 3 · α · L03 · Δ T
3 . 17 · 10– 6° c– 1 · (0.25 m)3 ·12°C
9.6 · 10 – 6m3
ΔV = 9.6 · 10 – 6m3

You can see that the change is small in both cases but is definitely something measurable.

Practice problems of this concept can be found at: Temperature and Heat Practice Questions

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