6.7 gm of water at be divided in two parts such that when one part of mass ‘x’ gm is turned into ice at , it would release sufficient amount of heat to vapourize the other part.
Answers
Given info : 6.7 gm of water at be divided in two parts such that when one part of mass ‘x’ gm is turned into ice at , it would release sufficient amount of heat to vapourize the other part.
To find : The value of x is ..
solution : case 1 : x gm is turned into ice.
so, heat released by x gm water , H = mLf
= x gm × 80 cal/gm
[ Lf is latent heat of fusion, for ice = 80 cal/gm]
= 80x cal
case 2 : remaining (6.7 - x) gm is turned into steam.
so, heat gained by (6.7 - x) gm of steam, H = mLv
= (6.7 - x) gm × 540 cal/gm
[ Lv is latent heat of vaporisation , for steam = 540 cal/g]
= 540 (6.7 - x) cal
heat gained = heat released
⇒80x = 540(6.7 - x)
⇒8x = 54 × 6.7 - 54x
⇒62x = 54 × 6.7
⇒x = 54 × 6.7/62 = 5.8 g
Therefore the value of x is 5.8
Answer:
Given info : 6.7 gm of water at be divided in two parts such that when one part of mass ‘x’ gm is turned into ice at , it would release sufficient amount of heat to vapourize the other part.
To find : The value of x is ..
solution : case 1 : x gm is turned into ice.
so, heat released by x gm water , H = mLf
= x gm × 80 cal/gm
[ Lf is latent heat of fusion, for ice = 80 cal/gm]
= 80x cal
case 2 : remaining (6.7 - x) gm is turned into steam.
so, heat gained by (6.7 - x) gm of steam, H = mLv
= (6.7 - x) gm × 540 cal/gm
[ Lv is latent heat of vaporisation , for steam = 540 cal/g]
= 540 (6.7 - x) cal
heat gained = heat released
⇒80x = 540(6.7 - x)
⇒8x = 54 × 6.7 - 54x
⇒62x = 54 × 6.7
⇒x = 54 × 6.7/62 = 5.8 g
Therefore the value of x is 5.8