A large punch bowl holds 3.35 kg of lemonade (which is essentially water) at 20.0° C.

Question

A 0.40-kg ice cube at
−10.2°
C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.

° C

kg

Answer 1

 

Mass of water =3.35 Kg

Temperature of water = 20C

Mass of Ice =0.4 Kg

Temp of ice = -10.2C

Heat lost by water till it reaches 0C =MSt = 3.35*4.2*10³*20 J

Heat gained by ice till it reaches 0C = MSt=0.4*2.1*10³*10.2 J

Difference=A?=Difference of above two values

Mass of mixture =3.75Kg

Temperature of mixture?

For finding temperature of mixture

Equate A =3.75*4.2*10³ *T

Find T = Final temperature of mixture= Temperature of system 

Remember here that thermal specific heat of water and ice are not the same