Rate of Heat Flow

Newton's law of cooling: the rate of loss of heat of a body is directly proportional to the difference between its temperature and that of its surroundings, provided this difference is small (about $30^\circ$C or less):

$\frac{dQ}{dt}\propto(\theta-\theta_0)$

Determining the specific heat capacity of a liquid by the cooling method

Principle: equal volumes of the liquid and of water (whose specific heat is known), in identical calorimeters, cool from the same temperature to the same temperature through the same excess over surroundings, so they lose heat at the same rate under the same conditions.

Procedure:

1. Weigh the empty calorimeter (mass $m_c$, specific heat $s_c$). Fill it with the hot liquid (mass $m_l$, specific heat $s_l$) to a fixed level, place in a draught-free enclosure, insert a thermometer, and stir gently.
2. Record the time $t_1$ taken for the temperature to fall from, say, $\theta_1$ to $\theta_2$ (e.g. $60^\circ$C to $50^\circ$C), the surrounding temperature being constant.
3. Repeat with the same calorimeter filled with the same volume of water (mass $m_w$, specific heat $s_w$), over the same temperature range, taking time $t_2$.

Calculation: Heat lost in the same range is proportional to time (same rate for the same conditions and same surface):

$\text{liquid: } (m_cs_c+m_ls_l)(\theta_1-\theta_2)=k\,t_1$

$\text{water: } (m_cs_c+m_ws_w)(\theta_1-\theta_2)=k\,t_2$

Dividing,

$\frac{m_cs_c+m_ls_l}{m_cs_c+m_ws_w}=\frac{t_1}{t_2}\;\Rightarrow\; s_l=\frac{(m_cs_c+m_ws_w)\,t_1/t_2-m_cs_c}{m_l}$

All the quantities on the right are known, so $s_l$ is found.

Dew forms when the surface cools below the dew point so that water vapour in the air condenses on it.

On a clear night the ground and grass radiate heat freely into space and nothing sends it back, so they cool quickly below the dew point and the vapour in contact with them condenses as dew.

On a cloudy night the clouds absorb the outgoing radiation and radiate part of it back to the ground, so the ground stays warmer and does not fall to the dew point. Hence little or no dew forms.

By the Stefan-Boltzmann law, the net power radiated per unit area by a black body at $T_1$ in surroundings at $T_2$ is

$\frac PA=\sigma\left(T_1^4-T_2^4\right)$

$T_1=327+273=600$ K, $T_2=27+273=300$ K.

$T_1^4=1.296\times10^{11},\qquad T_2^4=8.1\times10^{9}$

$\frac PA=5.67\times10^{-8}\times(1.296\times10^{11}-0.081\times10^{11})=5.67\times10^{-8}\times1.215\times10^{11}\approx6.9\times10^{3}\ \text{W m}^{-2}$

The net power loss per unit area is about $6.9\times10^{3}\ \text{W m}^{-2}$ (about 6.9 kW per square metre).

Emissivity (relative emittance): the ratio of the power radiated per unit area by a surface to the power radiated per unit area by a perfect black body at the same temperature, $e=\dfrac{E}{E_b}$. It is dimensionless and lies between 0 and 1.

Calculation: By Stefan's law for a non-black body, the energy radiated is $Q=e\sigma AT^4t$.

$Q=0.85\times5.67\times10^{-8}\times4\times10^{-5}\times(5000)^4\times60$

Power: $(5000)^4=6.25\times10^{14}$, so $P=0.85\times5.67\times10^{-8}\times4\times10^{-5}\times6.25\times10^{14}\approx1.20\times10^{3}\ \text{W}$

Energy in one minute: $Q=1205\times60\approx7.2\times10^{4}\ \text{J}$

The filament radiates about $7.2\times10^{4}$ J per minute.

(a) Two thin blankets trap a layer of air between them. Air is a very poor conductor of heat, so this layer greatly cuts heat loss from the body. A single blanket of double the thickness has no such layer, so it is less warm.

(b) Using the conduction formula with the thermal conductivity of glass $k\approx1\ \text{W m}^{-1}\text{K}^{-1}$:

$\frac{Q}{t}=\frac{kA\Delta\theta}{d}=\frac{1\times2\times(20-5)}{3\times10^{-3}}=1\times10^{4}\ \text{W}$

So the estimated rate of heat loss is about $10^4$ W (10 kW). This is an upper estimate; in practice the thin films of still air on each side of the glass reduce the loss considerably.

(a) The temperature gradient is the rate of change of temperature with distance, measured along the direction of heat flow (perpendicular to the isothermal surfaces):

$\text{temperature gradient}=\frac{\Delta\theta}{\Delta x}$

Its SI unit is $\text{K m}^{-1}$.

(b) For steady conduction through a flat slab of thickness $d$,

$H=\frac{kA(\theta_1-\theta_2)}{d}$

Here $k=0.040\ \text{W m}^{-1}\text{K}^{-1}$, $A=1.40\ \text{m}^2$ and $\theta_1-\theta_2=175-35=140\ \text{K}$, so $H=\dfrac{7.84}{d}\ \text{W}$ with $d$ in metres. The thickness is not given in the question text; in the standard version of this problem the fibre glass is $4.0\ \text{cm}$ thick:

$H=\frac{0.040\times1.40\times140}{0.040}=196\ \text{W}$

(c) The rate of conduction increases. Dry fibre glass traps air, which is a very poor conductor. If moisture is absorbed, water replaces some of the air; water conducts heat far better than air, so the effective thermal conductivity $k$ of the insulation rises, and since $H\propto k$ the heat flow rises.

The temperature gradient is the rate of change of temperature with distance along the direction of heat flow, i.e. the temperature difference per unit length measured perpendicular to the isothermal surfaces, $\dfrac{d\theta}{dx}$. Its SI unit is $\text{K m}^{-1}$.

For steady conduction through a flat slab,

$H=\frac{kA(\theta_1-\theta_2)}{d}$

With $k=0.040\ \text{W m}^{-1}\text{K}^{-1}$, $A=1.40\ \text{m}^2$ and $\theta_1-\theta_2=175-35=140\ \text{K}$:

$H=\frac{0.040\times1.40\times140}{d}=\frac{7.84}{d}\ \text{W}\quad(d\text{ in m})$

The thickness is missing from the question text. In the standard version of the problem the fibre glass is $4.0\ \text{cm}$ thick, which gives

$H=\frac{7.84}{0.040}=196\ \text{W}$

The rate of conduction increases. Dry fibre glass traps air, a very poor conductor, which is what makes it a good insulator. If it absorbs moisture, water displaces the air in its pores. Water has a much higher thermal conductivity than air, so the effective conductivity of the fibre glass rises, and since $H=\dfrac{kA\,\Delta\theta}{d}$ is proportional to $k$, heat flows through faster.

Black body. A perfect black body is one that absorbs all the radiation incident on it, whatever the wavelength, and reflects or transmits none. When heated it also emits the maximum possible radiation at a given temperature.

Realization (Ferry's black body). A hollow metal sphere has its inner surface coated with lamp black and a small hole on one side, with a conical projection opposite the hole. Radiation entering through the hole is reflected repeatedly inside and is almost completely absorbed; very little can escape. The hole therefore acts as a nearly perfect black body. When the sphere is heated uniformly, radiation coming out of the hole is black body radiation.

Stefan's law. The total energy radiated per second per unit area of a black body is directly proportional to the fourth power of its absolute temperature:

$E=\sigma T^4$

where $\sigma=5.67\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$ is Stefan's constant. For a black body at temperature $T$ in surroundings at $T_0$, the net loss of energy per unit area per second is

$E_{\text{net}}=\sigma\,(T^4-T_0^4)$

For a non-black body, $E=e\sigma T^4$, where $e<1$ is the emissivity.