Question

A hot liquid contained in a container of negligible heat capacity loses temperature at rate 3K/min just before it begins to solidify. The temperature remains constant for 30 min. If the ratio of specific heat capacity of liquid to specific latent heat of fusion is $\dfrac{x}{90}{{K}^{-1}}$. Find x. (given that the rate of losing heat is constant).

Answer 1

Hint: The rate of cooling of the liquid will be $\dfrac{dQ}{dt}=mc\dfrac{dT}{dt}$. It is given that $\dfrac{dT}{dt}=3\dfrac{K}{\min }$. Hence, find the rate of cooling of the liquid. Then it is given that the liquid begins to solidify and completely solidifies in 30 min at the same rate of cooling. The rate of losing heat is $\dfrac{dQ}{dt}=\dfrac{mL}{30}$. Equate both the rate of heat loss equations and the ratio of c to L.

Formulas Used:
$\dfrac{dQ}{dt}=mc\dfrac{dT}{dt}$
$Q=mL$

Complete step by step answer:
When there is a change in temperature of a substance, it either gives out heat (loses heat) or takes in heat (absorbs heat) depending whether the substance is cooling or heating.
When the temperature of a substance rises, it absorbs heat. When the temperature of the substance is lowered, it loses heat.
The heat gain or heat loss for a change of temperature of $\Delta T$ of the substance is given as $Q=mc\Delta T$.
Here, m is the mass of the substance and c is the specific heat of the substance.
The rate of cooling or heating with respect to time is given as $\dfrac{dQ}{dt}=mc\dfrac{dT}{dt}$, where $\dfrac{dT}{dt}$ is the rate of change of temperature.
In the given case, the liquid is cooling at a constant rate. It is given that the liquid loses temperature at rate 3K/min.
This means that $\dfrac{dT}{dt}=3\dfrac{K}{\min }$.
Therefore, rate of the cooling of the liquid is $\dfrac{dQ}{dt}=mc\dfrac{dT}{dt}=mc\left( 3 \right)=3mc$ ….. (i)
It is given that the liquid cools and begins to solidify. This means that the liquid undergoes phase transformation. When a substance undergoes phase transformation, its temperature remains constant.
If the phase transformation is from liquid to solid then the substance loses heat.
The heat lost by the substance is called as latent heat of fusion and is given as $Q=mL$, L is the specific latent heat of fusion.
It is given that the rate of losing heat is constant. It is given that the temperature remains constant for 30mins. This means that the liquid takes 30 min to completely solidify. And for this it loses a heat equal to mL.
Therefore, the rate of losing heat is $\dfrac{dQ}{dt}=\dfrac{mL}{30}$ … (ii).
Rate of cooling is also the rate of losing heat. Therefore, $\dfrac{dQ}{dt}=\dfrac{mL}{30}=3mc$.
$\Rightarrow \dfrac{mL}{30}=3mc$
$\Rightarrow \dfrac{c}{L}=\dfrac{1}{90}{{K}^{-1}}$.
This means that the ratio of specific heat capacity of liquid to specific latent heat of fusion is $\dfrac{1}{90}{{K}^{-1}}$.

Note:
We can also find the rate of cooling in the following way.
The heat loss for a change of temperature $\Delta T$ is given as $Q=mc\Delta T$.
The rate of cooling is equal to the heat lost in one unit of time (in this case it is one minute). Therefore, the rate of cooling will be the heat lost by the liquid in one minute.
It is given that in one minute the temperature changes by 3K. Therefore, the heat lost in one minute will be 3mc. Hence, the rate of cooling is equal to 3mc.