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Question: 3gm of wet steam (a mixture of steam and boiling water) is passed into \( 70.0{\text{ g}} \) of wate...

3gm of wet steam (a mixture of steam and boiling water) is passed into 70.0 g70.0{\text{ g}} of water at 12o C{\text{1}}{{\text{2}}^{\text{o}}}{\text{ C}} inside a calorimeter of mass 120.0 g120.0{\text{ g}} and specific heat capacity 420 J kg−1420{\text{ J k}}{{\text{g}}^{ - 1}} . Find the mass of water carried over with the steam if temperature is 32o C{\text{3}}{{\text{2}}^{\text{o}}}{\text{ C}} .
Specific latent heat of vaporization =2.3×106 J kg−1= 2.3 \times {10^6}{\text{ J k}}{{\text{g}}^{ - 1}} .
Specific heat capacity of water =4200 J kg−1 K−1= 4200{\text{ J k}}{{\text{g}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}

Explanation

Solution

To answer this question, you must recall the concept of bomb calorimetry. Bomb calorimetry, also known as constant volume calorimetry is a process used for measuring the heat of a reaction while maintaining constant volume and resisting high pressure. We shall assume the amount of water as boiling water and steam and substitute the values in the formula given.

Formula used: q=m×c×ΔTq = m \times c \times \Delta T
Where, qq represents the heat lost or gained by the substance
mm represents the mass of the substance
cc represents the specific heat of the substance
And, ΔT\Delta T represents the change in the temperature occurring.

Complete step by step solution:
Since we know that energy can neither be created nor be destroyed, we can conclude that in a closed system like a calorimeter, the energy of the system inside remains constant.
We are given 3.0 g3.0{\text{ g}} of wet steam. Let the amount of water in it be x and the amount of steam in it be y. Since we are given boiling water, the temperature can be taken as 100oC{\text{10}}{{\text{0}}^{\text{o}}}{\text{C}} considering atmospheric pressure. We can write, 3=x+y3 = {\text{x}} + {\text{y}}
Also since there will be no change in the overall energy of the system, we can write
Heat loss of wet steam = Heat gained by water + Heat gained by Calorimeter
We know that the formula for calculating the heat lost and gained is given by q=m×c×ΔTq = m \times c \times \Delta T
Also we are given the specific heat capacity of the calorimeter and water, so we can write:x×2.3×106+4200x(100−32)+y×4200(100−32)=70×4200(32−12)+120×420(32−12)x \times 2.3 \times {10^6} + 4200x\left( {100 - 32} \right) + y \times 4200\left( {100 - 32} \right) = 70 \times 4200\left( {32 - 12} \right) + 120 \times 420\left( {32 - 12} \right)
⇒2.3x×106+(x+y)4200(68)=84×4200×20\Rightarrow 2.3x \times {10^6} + \left( {x + y} \right)4200\left( {68} \right) = 84 \times 4200 \times 20
⇒2.3x×106+(3)4200(68)=84×4200×20\Rightarrow 2.3x \times {10^6} + \left( 3 \right)4200\left( {68} \right) = 84 \times 4200 \times 20
Solving, we get, x=2.7 gx = 2.7{\text{ g}} and thus, y=0.3 gy = 0.3{\text{ g}} .

Note:
The first law of thermodynamics is a modified version of the law of conservation of energy for thermodynamic processes. Its basic principle in the conservation of energy, i.e. energy can neither be created nor be destroyed in a thermodynamic process.