关于热传导的示例问题
热量可以通过任何由原子和分子组成的物质传递。在任何时候，原子都处于多种运动状态。热量或热能是由分子和原子的运动产生的，它存在于所有物质中。
分子移动越多，释放的热能就越多。然而，当谈到传热时，它只是指将热量从高温体传递到低温体的行为。热量可能以多种方式从一个位置转移到另一个位置。同时，如果两个系统有温差，热量就会找到一种方法从上层系统流到下层系统。
以下是传热方式：传导、对流和辐射。
什么是传导？
The process of heat transfer from things with higher temperatures to items with lower temperatures is known as conduction.
热能从高动能区转移到低动能区。当高速粒子与慢速粒子碰撞时，慢速粒子的动能增加。传导可以发生在固体、液体和气体中。
当热量通过传导从一个分子传递到另一个分子时，热能通常从一个分子传递到另一个分子，因为它们是直接接触的。然而，分子的位置保持不变。他们只是彼此产生共鸣。
传导方程
在传导方面，导热系数表明金属体更好地传递热量。
以下等式可用于计算传导率：
q = Q ⁄ t = K A (Th – Tc) ⁄ d
where K is the thermal conductivity, q is heat transfer rate, t is the time of transfer, Q is the amount of heat transfer,A is the area of surfaced is the thickness of the body, Th is the temperature of the hot region and Tc is the temperature of the cold region.
示例问题
问题 1：一块 10 厘米厚的冰块，温度为 0°C，位于 2400 厘米2石板的上表面。板坯在 100 °C 的温度下暴露在下表面上。假设冰的熔化潜热为 80 cal ⁄ gm，则在一小时内融化 4000 g 冰，求石头的热导率。
解决方案：
Given:
Area of slab, A = 2400 cm2
Thickness of ice, d = 10 cm
Temperature difference, Th – Tc = 100 °C – 0 °C = 100 °C
Time of heat transfer, t = 1 hr = 3600 s
Amount of heat transfer, Q = m L = 4000 × 80 = 320000 cal
Heat transfer rate, q = Q ⁄ t = 320000 cal ⁄ 3600 s = 89 cal ⁄ s
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of K.
K = q d ⁄ A (Th – Tc)
= (89 × 10) ⁄ (2400 × 100) cal ⁄ cm s °C
= 3.7 × 10-3 cal ⁄ cm s °C
Hence, the thermal conductivity of stone is 3.7 × 10-3 cal ⁄ cm s °C.
问题 2：一根 0.4 m 长和 0.04 m 直径的金属棒，一端为 373 K，另一端为 273 K。计算 1 分钟内传导的总热量。 （给定 K = 385 J ⁄ ms °C）
解决方案：
Given:
Thermal conductivity, K = 385 J ⁄ m s °C
Length of rod, d = 0.4 m
Diameter of rod, D = 0.04 m
Area of slab, A = π D2 ⁄ 4 = 0.001256 m2
Temperature difference, Th – Tc = 373 K – 273 K = 100 K
Time of heat transfer, t = 1 min = 60 s
The formula for heat transfer rate is given as:
Q ⁄ t = K A (Th – Tc) ⁄ d
Q = K A t (Th – Tc) ⁄ d
= (385 × 0.001256 × 60 × 100) ⁄ 0.4 J
= 7.25 × 103 J
Hence, the total amount of heat transfer is 7.25 × 103 J.
问题3：将等长2.0m、截面积2cm 2的铝棒和铜棒平行焊接在一起。一端保持在 10 °C，另一端保持在 30 °C。计算每秒从热端带走的热量。 （铝的导热系数为 200 W ⁄ m °C，铜的导热系数为 390 W ⁄ m °C）。
解决方案：
Given:
Thermal conductivity of aluminium, KAl = 200 W ⁄ m °C
Thermal conductivity of aluminium, KCu = 390 W ⁄ m °C
Combined thermal conductivity for parallel combination, K = 200 W ⁄ m °C + 390 W ⁄ m °C = 590 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2 = 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
= (590 × 2 × 10-4 × 20) ⁄ 2 W
= 1.18 W
Hence, the total amount of heat transfer is 1.18 W.
问题4：能量通过地表向外传导的平均速率为50.0 mW ⁄ m 2 ，近地表岩石的平均热导率为2.00 W ⁄ m K。假设地表温度为20.0° C，求 25.0 km 深度处的温度。
解决方案：
Given:
Average thermal conductivity, K = 2.00 W ⁄ m K
Depth, d = 25.0 km = 2.50 × 104 m
Surface temperature, Tc = 20.0 °C = (20 + 273) K = 293 K
Heat transfer rate per unit area, q ⁄ A = 50.0 mW ⁄ m2 = 50.0 × 10-3 W ⁄ m2
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of Th.
Th = q d ⁄ KA + Tc
= ((50.0 × 10-3 × 2.00 × 104) ⁄ 2.00) + 293
= (500 + 293) K
= 893 – 273 K
= 520 °C
Hence, the temperature at depth of 25.0 km is 520 °C.
问题 5：一块 10 cm 厚的钢板损失的能量为 50 W。假设温度差为 10.0 K，求钢板的面积。 （钢的热导率 = 45 W ⁄ m K）。
解决方案：
Given:
Thermal conductivity, K = 45 W ⁄ m K
Thickness of slab, d = 10 cm = 0.1 m
Temperature difference, Th – Tc = 10.0 K
Energy lost per sec, q = 50 W
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of A.
A = q d ⁄ K (Th – Tc)
= (50 × 0.1) ⁄ (45 × 10.0) m2
= 0.011 m2
Hence, the area of the slab is 0.011 m2.
问题 6：边长为 5 米的铝立方体的一个面保持在 60 ºC，另一端保持在 0 ºC。所有其他表面都被绝热壁覆盖。求 2 秒内流过立方体的热量。 （铝的热导率为 209 W ⁄ m ºC）。
解决方案：
Given:
Edge length of cube, d = 5 m
Surface area of cube, A = d2 = (5 m)2 = 25 m2
Temperature difference, Th – Tc = 60 ºC – 0 ºC = 60 ºC
Thermal conductivity, K = 209 W ⁄ m ºC
Heat transfer time, t = 2 sec
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
= (209 × 25 × 60) ⁄ 5 J
= 62700 J
= 62.7 KJ
Hence, the amount of heat that flows through the cube is 62.7 KJ.
习题7：等长2.0m、截面积2cm 2的铝棒和铜棒串联焊接在一起。一端保持在 10 °C，另一端保持在 30 °C。计算每秒从热端带走的热量。 （铝的导热系数为 200 W ⁄ m °C，铜的导热系数为 390 W ⁄ m °C）。
解决方案：
Given:
Thermal conductivity of aluminium, KAl = 200 W ⁄ m °C
Thermal conductivity of aluminium, KCu = 390 W ⁄ m °C
Combined thermal conductivity for parallel combination, 1 ⁄ K = 1 ⁄ 200 W ⁄ m °C + 1 ⁄ 390 W ⁄ m °C
K = (200 × 390) ⁄ (200 + 390) W ⁄ m °C
= 132.2 W ⁄ m °C
Length of rod, d = 2 m
Area of rod, A = 2 cm2 = 2 × 10-4 m2
Temperature difference, Th – Tc = 30 °C – 10 °C = 20 °C
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
= (132.2 × 2 × 10-4 × 20) ⁄ 2 W
= 0.2644 W
Hence, the total amount of heat transfer is 0.2644 W.