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ICSE Class 10 Answered

a vessel of negligile heat capacity contains 40gms of ice at 0°c. 8gms of steam at 100°c is passed into the ice to melt it. find the final temperature of the contents. ['L' of ice =336J/kg;'L' of steam =2268J/gm;'c' of water =4.2J/gm °c]
Asked by sonubharadwaj2000 | 12 Nov, 2018, 03:56: PM
answered-by-expert Expert Answer
(Latent heat of ice = 336 J / g  not 336 J/kg)
 
Let T be the final temperature
 
Heat loss for steam = 8×2268 + 8×4.2×(100-T) .....................(1)
Heat gained by ice  = 40×336 + 40×4.2×T ........................(2)
 
By energy conservation, heat loss = heat gain
hence by equating (1) and (2), we can solve for final temperature T. we get T = 40 °C
Answered by Thiyagarajan K | 12 Nov, 2018, 10:28: PM

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