Question

if ice at o°c and 40gm of mass is supplied 5000cal find the ginal temperature​

Answer 1

Answer:

Explanation:

q=ml

5000=40(delt)

5000/40=del t

125=del t

125=t2-t1

125=t2-0

125=t2

Answer 2

Answer:

Let t be the final temperature.

Then heat Liberated by water = m x c x Δt

= [200 x 10-3 x 4.2 x 103 x (50 – t)]

= (42,000 – 840t)

Heat absorbed by ice to change into water at 0oC = 40 x 10 x 336 x 103 = 13,440 J.

Heat absorbed by water to change its temperature from 0oC to toC = (40 x 10-3 x 4.2x 103 x t) = 168t

Total heat absorbed by water = (13,440 + 168t) J

According to the principle of calorimetry,

Heat given = Heat taken

or 42000 – 840t = 13440 + 168t

or 42000 – 13440 = 168t + 840t

or 1008t = 28560

or t = 28560/1008 = 28.33oC

Hence, the final temperature of water is 28.33oC