Some properties of a certain glass are listed here:
Density $2300 \mathrm {~kg} / \mathrm { m } ^ { 3 }$
Specific heat capacity $\quad 840 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$
Coefficient of thermal expansion $8.5 \times 10 ^ { - 6 } \left( \mathrm { C } ^ { \circ } \right) ^ { - 1 }$
Thermal conductivity $\quad 0.80 \mathrm {~W} / \mathrm { m } \cdot \mathrm { C } ^ { \circ }$
A glass window pane is 2.7 m high, 2.4 m wide, and 9.0 mm thick. The temperature at the inner surface of the glass is $19 ^ { \circ } \mathrm { C }$ and at the outer surface $4 ^ { \circ } \mathrm { C }$ How much heat is lost each hour through the window?

Question

Some properties of a certain glass are listed here:
Density $2300 \mathrm {~kg} / \mathrm { m } ^ { 3 }$
Specific heat capacity $\quad 840 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { C } ^ { \circ }$
Coefficient of thermal expansion $8.5 \times 10 ^ { - 6 } \left( \mathrm { C } ^ { \circ } \right) ^ { - 1 }$
Thermal conductivity $\quad 0.80 \mathrm {~W} / \mathrm { m } \cdot \mathrm { C } ^ { \circ }$
A glass window pane is  2.7 m high,  2.4 m wide, and  9.0 mm thick. The temperature at the inner surface of the glass is $19 ^ { \circ } \mathrm { C }$ and at the outer surface $4 ^ { \circ } \mathrm { C }$ How much heat is lost each hour through the window?

Quizplus · Accepted Answer

The heat loss can be calculated using the formula $ Q = \frac{k \cdot A \cdot \Delta T \cdot t}{d} $, where $ k $ is the thermal conductivity, $ A $ is the area, $ \Delta T $ is the temperature difference, $ t $ is the time, and $ d $ is the thickness. Substituting the given values, the heat loss is approximately $ 3.1 \times 10^7 $ J.

Some Properties of a Certain Glass Are Listed Here: Density 2300 kg/m32300 \mathrm {~kg} / \mathrm { m } ^ { 3 }2300 kg/m3

Some Properties of a Certain Glass Are Listed Here:
Density $2300 \mathrm {~kg} / \mathrm { m } ^ { 3 }$