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Answers
It is given that,
Mass = 100 g
Latent heat = 540 cal/g
Final temperature = 100°C
Initial temperature = 20°C
Heat rejected during Condensation of steam = Mass x Latent heat
= 100 x 540
= 54 x 10^3 cal. ------------- 1.
Heat rejected during Cooling = 100 x 1(100-20)
= 100 x 80 cal.
= 8 x 10^3 cal. ------------ 2
From eq (1) & (2)
Total heat rejected = (54 x 10^3 + 8 x 10^3) cal/min
= 62 x 10^3 cal/min.
= 6.2 x 10^4 cal/min
Be Brainly
Given,
a steam engine intakes 100 g of steam at 100° C per minute and cools it down to 20° C.
While cooling it down, steam engine will reject heat in two terms. First when temperature of steam got down to 100° C to 20°C and second, when steam changes into water(Liquid form)
1. Calculate heat rejected in cooling down steam at 100° C to 20°C ,
Q₁ = mass × specific heat of water × (temperature diff.)
Q₁ = 100 × 1 × (100-20)
Q₁ = 100 × 80
Q₁ = 8 × 10³ cal.
Note: Specific heat of water = 1 cal/g-°C
2. Calculate heat rejected in condensation of steam.
Q₂ = mass × latent heat
Q₂ = 100 × 540
Q₂ = 54 × 10³ cal.
Total heat rejected = Q₁ + Q₂
= 8 × 10³ + 54 × 10³
= 62 × 10³
= 6.2 × 10⁴ cal./per minute