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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 30min . Find ratio of specific heat capacity of liquid to specific latent heat of fusion in K⁻¹ (given that rate of losing heat is constant)

User Denski
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Final answer:

The ratio of the specific heat capacity of a liquid to its specific latent heat of fusion is 1/90 K⁻¹, based on the given rate of cooling before solidification and the time it takes for the temperature to remain constant during solidification.

Step-by-step explanation:

The question asks us to find the ratio of the specific heat capacity of a liquid to its specific latent heat of fusion in K⁻¹, using information about the rate of cooling and time taken for the phase change. Before solidification, the liquid cools at a constant rate of 3K/min, which implies a uniform rate of heat loss. When the liquid begins to solidify, the temperature remains constant for 30 minutes. This period of constant temperature indicates that heat is being used for the phase change at the liquid's freezing point instead of changing the temperature.

We can use the provided information to calculate the ratio.

Firstly, the amount of heat lost per minute (Q_loss) while cooling is represented by the formula:

Q_loss = mcΔT, where m is the mass of the substance, c is the specific heat capacity of the liquid, and ΔT is the change in temperature per minute.

For the phase change, the heat used to solidify the liquid (Q_fusion) is:

Q_fusion = mL, where L is the latent heat of fusion.

Since the heat loss rate is constant, we can equate the heat lost per minute before solidification (Q_loss) to the heat used to solidify the liquid per minute during the phase change (Q_fusion/30).

By setting these equal and canceling the mass (assuming it remains the same), we get:

mcΔT = L/30

Therefore, the ratio of specific heat capacity c to specific latent heat L is:

c/L = (1/(30ΔT))

Since ΔT is 3K, the ratio is:

c/L = 1/(30×3) in K⁻¹

c/L = 1/90K⁻¹

User Subodhkalika
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