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What is the final temperature of water?
701 cal of heat is added to 5.00 g ice at -20.0 celsius.
What is the final temperature of the water in celsius?
1 Answer
- NikhileshLv 78 years agoFavorite Answer
specific heat of ice = 2.09 J/g·°C
..
Latent heat of ice is 334.96 j/gm
Mass of ice is 5 gm , heat supplied 701*4.187 j = 2935 j
Heat supplied = 2935 = mass of ice * sp heat of ice * ( 0- (-20) ) + mass of ice * latent heat of ice + 5*1* (t -0)
Heat supplied would first raise temp of ice to zero degc, at zero degc ice melts to water and then water temp would rise depending on avl heat
or ,2935 = 5*2.09*20 + 5*334.96 +5(t -0)
2935 = 1884 + 5 ( t-0)
5 ( t - 0) = 210
t = 42.04
