传热公式
热量是可以从一个点传递到另一个点的热能的量度。热是动能从能源到介质或从一种介质或物体到另一种介质或物体的传递。
热量是与功和能量相关的相变的重要组成部分之一。热量也是系统中粒子所拥有的动能的量度。系统中粒子的动能随着系统温度的升高而增加。因此，热量测量随时间而变化。
传播热量
当较高温度的系统与较低温度的系统接触时，能量从第一个系统中的粒子转移到第二个系统中的粒子。因此，传热可以定义为热量从温度较高的物体（或系统）传递到温度较低的另一个物体（或系统）的过程。
传热公式
传热公式确定从一个系统传递到另一个系统的热量。
Q = c × m × ΔT
Where,
Q is the heat supplied to the system
m is the mass of the system
c is the specific heat capacity of the system
ΔT is the change in temperature of the system
比热容 (c) 定义为每单位质量 (kg) 的材料在温度增加 1 K（或 1 °C）时所吸收的热量（以焦耳为单位）。其单位为 J/kg/K 或 J/kg/°C。
公式的推导
令m为系统的质量， c为系统的比热容。令ΔT为系统的温度变化。
那么所提供的热量 ( Q ) 是质量m 、比热容c和温度变化ΔT的乘积，由下式给出，
Q = c × m × ΔT
传热类型
传热方式分为三种：
- 传导
- 对流
- 辐射
传导
通过固体材料的热传递称为传导。通过传导过程传递热量的公式表示为：
Q = kA(THot-TCold)t/d
Where,
Q is heat transferred through conduction
k is thermal conductivity of the material
A is the area of the surface
THot is the temperature of the hot surface
TCold is the temperature of the cold surface
t is time
d is the thickness of the material
对流
通过液体和气体进行的热传递称为对流。对流过程传热的公式表示为：
Q = HcA(THot-TCold)
Where,
Q is heat transferred through convection
Hc is the heat transfer coefficient
A is the area of the surface
THot is the temperature of the hot system
TCold is the temperature of the cold system
辐射
通过电磁波传递热量称为辐射。辐射过程传递热量的公式表示为：
Q = σ (THot – TCold)4A
Where,
Q is heat transferred through radiation
σ is Stefan Boltzmann Constant
THot is the temperature of the hot system
TCold is the temperature of the cold system
A is the area of the surface
Stefan Boltzmann 常数 (σ) 计算如下：
σ = 2.π5 KB4 / 15 h3 c2 = 5.670367(13) × 10-8 J . m-2. S-1 . K-4
Where,
σ is Stefan Boltzmann Constant
pi(π) ∼= 3.14
kB is Boltzmann constant
h is Planck’s constant
c is speed of light in vacuum
示例问题
问题1：将一个质量为10 kg，初始温度为200 K 的系统加热到450 K。系统的比热容为0.91 KJ/kg K。计算系统在此过程中获得的热量。
解决方案：
According to question,
Mass, m = 10 kg
Specific heat capacity, c = 0.91 KJ/kg K
Initial temperature, Ti = 200 K
Final temperature, Tf = 450 K
Change in temperature, ΔT = 450K – 200K = 250K
Using the heat transfer formula,
Q = c × m × ΔT
Q = 0.91 x 10 x 250
Q = 2275 KJ
Therefore the total heat gained by the system is 2275 KJ.
问题2：铁的比热为0.45 J/g°C。如果温度变化为 40°C，1200 焦耳的热传递需要多少铁？
解决方案：
According to question,
Specific heat of iron, c = 0.45 J/g°C
Change in temperature, ΔT = 40°C
Amount of heat transferred, Q = 1200 J
Using the heat transfer formula,
Q = c × m × ΔT
m = Q /(c x ΔT)
m = 1200 /(0.45 x 40)
m = 66.667 g
Therefore required mass of iron for a heat transfer of 1200 Joules is 66.667 grams.
问题3：考虑由长3m、宽1.5m、厚0.005m的玻璃墙隔开的两个不同温度的水柱。一个水柱为 380K，另一个为 120K。如果玻璃的热导率为 1.4 W/mK，则计算传递的热量。
解决方案：
According to question,
Thermal Conductivity of glass, k = 1.4 W/mK.
Temperature of first water column, THot= 380K
Temperature of second water column, TCold = 120K
Area of the glass wall separating two columns, A = length x width = 3m x 1.5m = 4.5m2
Thickness of the glass, d = 0.005m
Using the heat transfer formula for conduction,
Q = kA(THot-TCold)t / d
Q = 1.4 x 4.5 (380-120) / 0.005
Q = 327600 W
Therefore, amount of heat transferred is 327600 Watts.
问题4：如果介质的传热系数为8 W/(m 2 K )，面积为25 m 2 ，温差为20K，计算对流传热。
解决方案：
According to question,
Heat transfer coefficient, Hc = 8 W/(m2 K)
Area, A = 25m2
Change in temperature, (THot – TCold) = 20K
Using the heat transfer formula for convection,
Q = HcA(THot-TCold)
Q = 8 x 25 x 20
Q = 4000 W
Therefore, amount of heat transferred through convection is 4000 Watts.
问题 5：计算温度为 300K 和 430K 的两个黑体之间通过辐射传递的热量，介质面积为 48 m 2 。 （给定 Stefan Boltzmann 常数，σ = 5.67 x 10 -8 W/(m 2 K 4 ) ）。
解决方案：
According to question,
Temperature of hot body, THot= 430K
Temperature of cold body, TCold = 300K
Change in temperature, (THot – TCold) = 430K – 300K = 130K
Area, A = 48 m2
Stefan Boltzmann Constant, σ = 5.67 x 10-8 W/(m2K4)
Using the heat transfer formula for radiation,
Q = σ (THot-TCold)4 A
Q = 5.67 x 10-8 x 1304 x 48
Q = 777.3 W
Therefore, amount of heat transferred through radiation is 777.3 Watts.