6ne replies????
Q1. In a container of negligible mass 140 g of ice initially at -15°C is added to 200g of water that has a temperature of 40°C.If no heat is lost to the surroundings ,what is the final temperature of the system and masses of water and ice in the mixture.
Q2.Equal masses of ice(at 0°C) and water are in contact .Find the temperature of water just needed to just melt the complete ice.
ANS-(1)0°C,mass of ice is 54g and that of water is 286g
(2)80°C
Note-I have just started thermal physics,so please dxplain with full steps.THANX.
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6 Answers
1first we bring the ice to 0°C and calculate how much heat is required to do so. let this be equal to H1
calculate the heat released when the water comes to 0°C
let this be equal to H2
if H1=H2 then the ice will not melt
if H2>H1 the ice will melt
let the mass of ice be M
let m mass of ice melts
Heat required to melt it = qm
where q is the latent heat of melting
then
qm=H2-H1
this will give u q
so the final mass of ice=M-m
mass of water =initial mass of water + m
1for 2nd
mass of ice*latent heat=masss of water*specific heat capacity*temp
n masss of both r same