Engineering Fundamentals: An Introduction to Engineering, 4th ed.c

334 Chapter 11 Temperature and Temperature-Related Parameters

change, the relationship among the required thermal energy (Ethermal), mass of the given mate-

rial (m), its specific heat (c), and the temperature rise (TfinalTinitial) that will occur is given by

(11.28)

where

Ethermalthermal energy ( J or Btu)

mmass (kg or lbm)

TfinalTinitialtemperature rise (°C or K, °F or °R)

Next we will look at two examples that demonstrate the use of Equation (11.28).

Example 11.15 An aluminum circular disk with a diameter,d, of 15 cm and a thickness of 4 mm is exposed to

a heat source that puts out 200 J every second. The density of the aluminum is 2700 kg /m

3

.

Assuming no heat loss to the surrounding, estimate the temperature rise of the disk after 15 s.

We will make use of Equation (11.28) and Table 11.11 to solve this problem, but first we

need to calculate the mass of the disk using the information given.

(every second)

And after 15 s the temperature rise will be 18C or 18 K.

Example 11.16 We have exposed 1 kg of water, 1 kg of brick, and 1 kg of concrete, each to a heat source that

puts out 100 J every second. Assuming that all of the supplied energy goes to each material and

they were all initially at the same temperature, which one of these materials will have a greater

temperature rise after 10 s?

We can answer this question using Equation (11.28) and Table 11.11. We will first look up

the values of the specific heat for water, brick, and concrete, which arecwater4180 J/ kgK,

cbrick960 J/ kgK, andcconcrete880 J/ kgK. Now applying Equation (11.28),Ethermal

mc(TfinalTinitial) to each situation, it should be clear that although each material has the same

amount of mass and is exposed to the same amount of thermal energy, the concrete will expe-

rience a higher temperature rise because it has the lowest heat capacity value among the three

given materials.

1 TfinalTinitial 2 1.2 K

200 J 1 0.191 kg 21 875 J/kg#K 21 TfinalTinitial 2

Ethermalmc 1 TfinalTinitial 2

m 1 7.06858 10

 5

m

3

21 2700 kg/m

3

2 0.191 kg

VolumeV

p

4

d

2

1 thickness 2 

p

4

1 0.15 m 2

2

1 0.004 m 2 7.06858 10

 5

m

3

massm 1 density 2 1 volume 2

cspecific heat 1 J/kg#K or J/kg#°C or Btu/lbm#°R or Btu/lbm#°F 2

Ethermalmc 1 TfinalTinitial 2

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