
A liquid at ${30^ \circ }C$ is poured very slowly into a Calorimeter that is at temperature of ${110^ \circ }C$. The boiling temperature of the liquid is ${80^ \circ }C$. It is found that the first 5gm of the liquid completely evaporates. After pouring another 80 gm of the liquid the equilibrium temperature is found to be ${50^ \circ }C$. The ratio of the Latent heat of the liquid to its specific heat will be ______ $^ \circ C$. (Neglect the heat exchange with surrounding).
Answer
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Hint
The calorimeter is defined to be an object that is used to measure the heat of any chemical reactions or the physical changes as well as the heat capacity. For an instance, if there is an occurrence of the exothermic reaction taking place in a solution in a calorimeter, then the heat produced by the reaction is absorbed by the solution, which will increase its temperature. Based on this concept we have to solve this question.
Complete step-by step answer
In case 1 we have to evaluate that:
If calorimeter is open and after evaporation liquid escapes
$5 \times S \times 50 + 5L = W \times 30\;....(1)$
$80 \times S \times 20 = W \times 30\;....(2)$
$80 \times S \times 20 = 5 \times S \times 50 \times 5L$
$5L = 1350S$
$\dfrac{L}{S} = 270$
In case 2 we can evaluate that:
If calorimeter is closed (vapour not allowed to escape)
Heat gain = Heat loss
$5S(80 - 30) + 5L = W(110 - 80)$
$S$ is Specific heat of liquid,
$L$ is Latent heat of liquid,
$W$ is Water equivalent of a calorimeter.
$250S + 5L = W \times 30\;....(1)$
Now $80gm$ liquid is poured
Heat gain = Heat loss
Here, the final temperature is ${50^ \circ }C$
$80 \times S \times 20 = 5L + 5S \times 30 + W \times 30\;....(2)$
From (1) and (2)
$\dfrac{L}{S} = 120$
Hence, the answer is 120.
Note
We know that a specific heat capacity of a substance is defined as the heat capacity of a sample of the substance divided by the mass of the sample. On the other hand, latent heat is defined as the heat of energy that is absorbed or released during a phase change of a substance. The latent heat is related to the heat property called enthalpy.
The calorimeter is defined to be an object that is used to measure the heat of any chemical reactions or the physical changes as well as the heat capacity. For an instance, if there is an occurrence of the exothermic reaction taking place in a solution in a calorimeter, then the heat produced by the reaction is absorbed by the solution, which will increase its temperature. Based on this concept we have to solve this question.
Complete step-by step answer
In case 1 we have to evaluate that:
If calorimeter is open and after evaporation liquid escapes
$5 \times S \times 50 + 5L = W \times 30\;....(1)$
$80 \times S \times 20 = W \times 30\;....(2)$
$80 \times S \times 20 = 5 \times S \times 50 \times 5L$
$5L = 1350S$
$\dfrac{L}{S} = 270$
In case 2 we can evaluate that:
If calorimeter is closed (vapour not allowed to escape)
Heat gain = Heat loss
$5S(80 - 30) + 5L = W(110 - 80)$
$S$ is Specific heat of liquid,
$L$ is Latent heat of liquid,
$W$ is Water equivalent of a calorimeter.
$250S + 5L = W \times 30\;....(1)$
Now $80gm$ liquid is poured
Heat gain = Heat loss
Here, the final temperature is ${50^ \circ }C$
$80 \times S \times 20 = 5L + 5S \times 30 + W \times 30\;....(2)$
From (1) and (2)
$\dfrac{L}{S} = 120$
Hence, the answer is 120.
Note
We know that a specific heat capacity of a substance is defined as the heat capacity of a sample of the substance divided by the mass of the sample. On the other hand, latent heat is defined as the heat of energy that is absorbed or released during a phase change of a substance. The latent heat is related to the heat property called enthalpy.
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