What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100ºC to 80ºC? [the specific latent heat of steam is 2,260,000Jkg\(^{-1}\), specific heat capacity of water is 4200Jkg\(^{-1}\)K\(^{-1}\)]
4200J
2,260,OOOJ
9,040,000J
9,380,000J
Explanation
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The correct answer is C: 9,040,000 \, \text{J} .
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Step-by-step explanation:
We are comparing 4 kg of steam at 100°C with 4 kg of water at 100°C cooled to 80°C.
1. Heat released by 4 kg of water cooling from 100°C to 80°C
This is just sensible heat (no phase change):
Q_{\text{water}} = m c \Delta T
Q_{\text{water}} = 4 \times 4200 \times (100 - 80)
Q_{\text{water}} = 4 \times 4200 \times 20 = 336,000 \, \text{J}
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2. Heat released by 4 kg of steam cooling from 100°C to 80°C
First step: steam condenses to water at 100°C
Latent heat released:
Q_{\text{latent}} = m L = 4 \times 2,260,000
Q_{\text{latent}} = 9,040,000 \, \text{J}
Second step: condensed water cools from 100°C to 80°C
Same as before:
Q_{\text{sensible (condensed)}} = 4 \times 4200 \times 20 = 336,000 \, \text{J}
Total for steam:
Q_{\text{steam}} = 9,040,000 + 336,000 = 9,376,000 \, \text{J}
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3. Difference in heat given out
\Delta Q = Q_{\text{steam}} - Q_{\text{water}}
\Delta Q = (9,040,000 + 336,000) - 336,000
\Delta Q = 9,040,000 \, \text{J}
This is just the latent heat of condensation part, because both cases have the same sensible cooling part afterward.
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Final answer: C: 9,040,000 \, \text{J}
the reasoning is this we are to find the difference in heat when we cool 4kg of steam and water their masses are the same now if you cool steam from 100 to 80°c you would need to break latent heat of vaporization first at 100°c then cool it normally as water to 80°c
so total heat for cooling steam will be (mc∅ + ml) that is heat for cooling + heat of condensing but to cool water water is already formed so latent heat is not needed the only heat will be heat of cooling which is (mc∅) since they both have same mass and same specific heat capacity and we are asked to find the difference in heat we can compare them like this
(mc∅ + ml) - (mc∅)
that is total heat of cooling steam - total heat of cooling water
mc∅- mc∅ + ml (on removing brackets) then the difference should be ml on calculating 4× the latent heat you should get the answer
hope this helps
