Question

2kg water vapour 100°C is condensed to water at 40°C. The quantity of heat released is​

Answer 1

first of all, find amount of energy released to condense 2kg water vapour.

i.e., H = mLv.

where m is mass of water vapour and Lv is latent heat of vaporisation i.e., Lv = 2256 KJ/kg

so, H = 2kg × 2256 KJ/kg = 4512 kg

again, heat released to decrease the temperature of water from 100° to 40°C , H' = ms∆T

where s is specific heat of water i.e., s = 4.2 KJ/kg .°C , °C and ∆T is change in temperature. so, (100 - 40) = 60°C

now, H' = 2kg × 4.2KJ/kg°C × 60°C = 2 × 4.2 × 60 KJ= 8.4 × 60 KJ= 504 KJ

hence, total heat released = H + H'

= 4512 + 504 = 5016 KJ = 5.016 × 10^6 J

Answer 2

$\huge \color{green} \boxed{\colorbox{lightgreen}{Ur\:Answer:)}}$

Mass of the steam = 2kg

At 100℃

L = 540cal/g.

= 54000cal/kg.

steam water

T = 100℃ $\longrightarrow$ T = 100℃

$Q_1$ $\downarrow Q_2$

Water

T = 40

$Q_1$ = mL

$\Rightarrow$ 2 ×54000

$\Rightarrow$ 108000cal

△T = 100- 40

$\Rightarrow$ 60

∴ $Q_{ 12}$= ms△T

= 2×1×100×60

$\Rightarrow$ 12000cal.

$\red{\underline{\bold{Energy\:evolved\::}}}$

$\Rightarrow$ $Q_1+Q_1$

$\Rightarrow$ 12000+108000

$\Rightarrow$ 120000cal

$\Rightarrow \sf \red{1200kcal}.$

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~hope it helps ❤

$\tt\purple{@Manogna}$