Question

certain window glass 30 x 36 inches is 1/8 in thick. one side has a uniform temperature of 70° f and the second face of temperature of 10°F what is the temperature gradient?​

Answer 1

Answer:

The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C

and 3°C, respectively. If the thermal conductivity of the glass is 0.78 W/m·K, determine the

amount of heat loss through the glass over a period of 5 h. What would your answer be if

the glass were 1 cm thick? (Problem 1.56 from text.)

This is a basic conduction problem where Q& =kAΔT/L. The heat loss from the inside to the

outside is found by taking ΔT = 10o

C – 3o

C = 7o

C = 7 K, the inside to outside temperature

difference. The area, A, is (2 m)(2 m) = 4 m2

, and we are given the thermal conductivity, k =

0.78 W/m·K. We are asked to find the heat for two values of L: 0.5 cm = 0.005 m and 1 cm =

0.01 m. The total heat transferred over five hours is found by assuming that the temperatures are

constant over that period so that Q = Q& Δt, where Δt = 5 hr = 18,000 s. Combining these

equations and applying the data gives the following results for L = 0.5 cm and L = 1 cm.

( )( )

( ) MJ

J

MJ

W s

J

s

m

m K

m K

W

t

L

kA T Q Q t 78.6

10

1 18000

0.005

0.78 4 7

6

2

= ⋅ ⋅ Δ = Δ = Δ = &

( )( )

( ) MJ

J

MJ

W s

J

s

m

m K

m K

W

t

L

kA T Q Q t 39.3

10

1 18000

0.01

0.78 4 7

6

2

= ⋅ ⋅ Δ = Δ = Δ = &

The second answer can also be found by dividing the first answer by 2 to account for the doubling

of the thickness of the glass.

Explanation:

Answer 2

Given:

Thickness of a certain window glass 30 x 36 inches is 1/8 inch. One side has a uniform temperature of 70° f and the second face of temperature of 10°F.

To find:

Temperature gradient?

Calculation:

Temperature gradient is defined as the rate of change of temperature with respect to distance.

• In this case , temperature gradient is the ratio of temperature difference and thickness of glass.

• Let temperature gradient be T_(g):

$\therefore \: T_{g} = \dfrac{\Delta T}{d}$

$\implies \: T_{g} = \dfrac{(70 - 10)}{( \frac{1}{8} )}$

$\implies \: T_{g} = (70 - 10) \times 8$

$\implies \: T_{g} = 60 \times 8$

$\implies \: T_{g} = 480 \:^{ \circ}C /inch$

So, the temperature gradient is 480°C/inch.