Heat required to convert 2 g ice at -20°C into water at 20°C is (Specific heat of ice 0.5 cal g-1 °C-1, latent heat of fusion of ice 80 cal g )
-20°C वाली 2 g बर्फ को 20°C वाले जल में रूपान्तरित करने में आवश्यक ऊष्मा है (बर्फ
की विशिष्ट ऊष्मा 0.5 cal g-1 °C-1 है, बर्फ के संलयन की गुप्त ऊष्मा 80 calg)
A
220 joule
220 जूल
220cal
100cal
22cal

Heat required = 220 cal

Given

2 g ice at -20°C into water at 20°C

c ice =  0.5 cal g-1 °C-1

Lf = 80 cal/g

Required

Heat

Solution

The heat needed to raise the temperature

Q = m . c . Δt

1. heat to raise temperature from -20 °C to 0 °C

Q = 2 x 0.5 x 20

Q = 20 cal

2. phase change(ice to water)

Q = m.Lf

Q = 2 . 80 cal/g

Q = 160 cal

3. heat to raise temperature from 0 °C to 20 °C ( c water = 1 cal/g.C)

Q = 2 . 1 . 20

Q = 40 cal

Total heat = 220 cal

Explanation:

KOH +HBr = KBr + H2O

The equation is balanced since all the molecules are equal..