Question

There is a layer of ice 6cm thick over the surface of a pond. The temperature of the air is 283 kelvin. If the rate of heat loss energy from the water be 200 joule per square meter per second, find the value of the thermal conductivity of ice.

Answer 1

In the above given question we will use the concept of thermal conductivity. The thermal conductivity of a material measures its ability to conduct heat. As per the given question we will substitute the values of the given quantities.
Formula: Rate of heat transfer is equal to $\dfrac{{\partial Q}}{{\partial t}} = \dfrac{{KA\Delta T}}{t}$

Complete answer:
To find the thermal conductivity of the ice we will modify the above formula as,
\eqalign{ & K = \dfrac{{\partial Q}}{{\partial t}} \times \dfrac{t}{{A\Delta T}} \cr & \Rightarrow K = \dfrac{{Qt}}{{A\Delta T}} \cr}
where the K is the thermal conductivity of ice, Q is the rate of heat loss energy in unit time from the water ,A is the area of cross section, t is the thickness and $\Delta T$ is the temperature difference given as$= {T_2} - {T_1}$, here ${T_1}$is the initial temperature that is 273 kelvin and ${T_2}$is the temperature of the air given as 283 kelvin.
Thus, as per the data given in the question
$Q = 200$joule per square meter per second
$A = 1{m^2}$
T=0.06m
$\Delta T = (283 - 273)Kelvin$
So,
$\Rightarrow K = \dfrac{{Qt}}{{A\Delta T}}$
$\Rightarrow K = \dfrac{{200 \times 0.06}}{{1 \times (283 - 273)}}$
$\Rightarrow K = \dfrac{{200 \times 0.06}}{{1 \times 10}}$
$\therefore K = 1.2$ $\dfrac{W}{{mK}}$
Therefore, the value of the thermal conductivity is 1.2$\dfrac{W}{{mK}}$

Additional information:
The thermal conductivity of a material depends on the area of cross section and on the thickness of the material. The thermal property of the material is also known as the transport property. As it provides the relationship between heat and temperature. It also provides the rate at which heat is transferred.

Note:
In such kinds of questions we must remember the formula and the relation between the thermal conductivity and the rate of heat transferred. In such kinds of the question students may commit mistakes in considering the units of the quantities given. The unit of the thermal conductivity is $\dfrac{W}{{mK}}$ and instead of this they might consider $\dfrac{J}{{mK}}$ which is wrong.