Question

plzz see in image and answer fast plzz i have a exan

Answer 1

1. Mass=1gm
From 0°water=>0°ice
Latent heat of fusion=80cal/g
Q=Lm
Q=80*1
Heat liberated=80cal

2. 100°steam=>100°water=>0°water=>0°ice
Latent heat of vapourisation=540
Q=540*1=540cal
Specific heat=1
Q=ms∆t
Q=1*1*100=100
Latent heat of fusion=80
Q=1*80=80
Total heat liberated=540+100+80=720cal

Answer 2

Hey there!

(c) Given,
Mass of water = 1gm.

We need to fine the heat released or absorbed.

so, Q = Latent heat of fusion = 80 cal.

Mass = 1gm. so 1×80 = 80 cal.

So, 80 cal of energy is transformed when water at 0° freezes to Ice at 0°.

------------------------

(d)

Step 1 :- Heat energy released or absorbed when 1gm of steam at 100° converted into water at 100° = Latent heat of vaporization = 540 cal.

Step 2 :- Heat energy released or absorbed when water at 100° converted into water at 0°

Q = ms∆t
Mass = 1gm. Specific heat = 1. Change in temperature = 100°.

= 1 × 1 ×100 = 100cal.

Step 3 :- Heat energy released or absorbed when water at 0° converted into ice at 0° = Latent heat of fusion = 80cal.

Adding up these, 540+100+80 = 720 cal.

Therefore 720 cal is transformed when steam at 100° is converted into ice at 0°.


If you have any doubt.. you can ask me.
:)