Question

In the following equation calculate the value of heat (H).
1kg steam (200°C) – 1 kg water (100°C) + H (heat)
(1) 640 kcal
(2) 540 kcal
(3) 130 kcal
(4) 590 kcal​

Answer 1

Answer:

In the given equation following processes takes place.

Process 1 : Steam at 200

0

C loses out heat to change to steam at 100

0

C

Heat released in this case = mcΔT

where, c = specific heat of steam = 0.5cal/g

0

C

So, for step 1, heat released = (1000g)0.5cal/g

0

C(200−100)

0

C=50000cal

Process 2 : Steam at 100

0

C gets converted to water at 100

0

C

This process releases heat. Latent heat of vaporisation of water = 540cal/g

0

C

So, 1000 g of steam would release 540000 cal heat.

So, the total heat released in the mentioned equation is (50000+540000) cal = 590000 cal heat = H

Answer 2

Given info : The equation is given as

1 kg steam (200°C) = 1 kg water (100°C) + H (heat)

To find : the value of heat (H) is ...

solution : heat released by steam to decrease the temperature from 200°C to 100°C = ms∆T

= 1kg × 0.5 Cal/g/°C × (200°C - 100°C)

= 50 kCal

heat released by steam to change the phase into water at 100°C = mLv

= 1kg × 540 kCal/kg

= 540 kCal

now water gotten from LHS side has same energy as given water in RHS side.

so, total heat released by steam = 50 kCal + 540 kCal = 590 kCal

so, 590 kCal + 1 kg water (100°C) = 1 kg water (100°C) + H

⇒H = 590 kCal

Therefore the value of H is 590 kCal.

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