Question

Just before it begins to harden, a heated liquid enclosed in a container with little heat capacity loses temperature at a rate of 3K/min. 

For 30 minutes, the temperature remains constant. 

If x/90K^-1 is the ratio of liquid specific heat capacity to specific latent heat of fusion, 

Look for x. 

(Assuming that the rate of heat loss remains constant).

Answer 1

The heat gain or heat loss for a change of temperature of ΔT of the substance is given as Q=mcΔT.

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 dQ/dt=mcdT/dt, where 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 dT/dt = 3Kmin.

Therefore, rate of the cooling of the liquid is dQ/dt=mc dT/dt=mc(3)=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 dQ/dt=mL/30 … (ii).

Rate of cooling is also the rate of losing heat. Therefore, dQ/dt = mL/30 = 3mc.

⇒ mL/30 = 3mc

⇒ c/L = 1/90K−1

 

 

The value of x = 1.