Question

A container with $1{\rm{ }}kg$ of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. Assuming that the heat loss from the water to the surrounding is governed by Newton's law of cooling, the difference (in$^ \circ C$) in the temperature of water and the surroundings after a long time will be_____? (Ignore effect of the container, and take constant for Newton’s law of cooling = $0.001{s^{ - 1}}$ , Heat capacity of water = $4200J\;k{g^{ - 1}}{K^{ - 1}}$)

Answer 1

Hint : Keep in mind Newton’s law of cooling, which states that the temperature of the body changes at a rate proportional to the difference in temperature between the body and its surroundings. And also keep in mind about heating capacity. It is a physical property of matter defined as the amount of heat supplied to a given mass of a material to produce a unit change in it’s temperature.

Complete step-by-step solution:
Firstly we need to derive the expression for the rate of loss of heat to the surroundings
Stefan-Boltzmann law states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. This law applies only to blackbodies, theoretical surfaces that absorb all incident heat radiation.
It is given by ;
$R = \sigma \varepsilon {T^4}(\dfrac{W}{{{m^2}}})$
From this law we can conclude that
$R \propto {T^4}$
Let us assume heat radiated $R$to be ${E_b}$for easier conversion.
So we get,
${E_b} \propto {T^4}$
Replacing the proportionality symbol, we get
${E_b} = \sigma {T^4}$
Where $\sigma =$constant known as Stefan’s constant
Let $T$be the absolute temperature of a perfectly black body. Let ${T_0}$be the absolute temperature of the surroundings.
Heat radiated per unit area of a perfectly black body = $\sigma {T^4}$
Let $A$be the surface area of the perfectly black body. Then,
Heat lost by the body per unit time = $A\sigma {T^4}$
The net rate of loss of heat = $A\sigma {T^4} - A\sigma {T_0}^4$
= $A\sigma ({T^4} - {T_0}^4)$
Which gives us our expression;
$\dfrac{{dQ}}{{dT}} = eA\sigma ({T^4} - {T_0}^4)$
For small temperature change, the expression changes into;
$\dfrac{{dQ}}{{dT}} = e\sigma A{T^3}\Delta T$ ……(1)
We also know that,
$Q = mc\Delta T$ ……(2)
Where $Q$ is heat transferred, and $c$ is symbol of specific heat
Putting (2) in (1) we get
$\dfrac{{mc\Delta T}}{{dT}} = e\sigma A{T^3}\Delta T$
$\dfrac{{\Delta T}}{{dT}} = \dfrac{{e\sigma A{T^3}}}{{mc}}\Delta T$
($\dfrac{{e\sigma A{T^3}}}{{mc}}$) = constant for newton law of cooling
And from question, we know constant for newton law of cooling is 0.001
Putting values in equation we get,
$\dfrac{{e\sigma A{T^3}}}{{mc}}$= 0.001
$e\sigma A{T^3} = mc \times 0.001$
= $1 \times 4200 \times 0.001$
$e\sigma A{T^3} = 4.2$ ……(3)
From question we know,
$\dfrac{{dQ}}{{dt}} =$ Average energy per unit time per unit area received $\times$ effective area
$\dfrac{{dQ}}{{dt}} = 700 \times 0.05 = 35W$ ……(4)
Putting values of (3) and (4) in (1) we get,
$35 = 4.2\Delta T$
$\dfrac{{35}}{{4.2}} = \Delta T$
$\Delta T = 8.33$
So to conclude, the difference in temperature will be 8.33.

Note: Newton’s law of cooling tells us the rate at which an exposed body changes its temperature through radiation which is approximately proportional to the difference between the object’s temperature and its surroundings, given that the temperature difference is small. Newton’s Law of cooling is directly proportional to temperature difference between system and surrounding.