Question

A 25 g glass tumbler contains 420 mL of water at 24°C. If four 12 g ice cubes each at a temperature of -3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room. 21.3

Answer 1

We know that latent heat of fusion of ice is 334 j/g

And specific heat of ice is 2.108 J/gK

Now,

The energy required for ice to reach at 0 ⁰C water is,

=(2.108*3*12)+(12*334)

=4083.88 J

Now, if water is coming to 0 ⁰C

then, energy released is

=420*4.18*24

=42134.4 J, So ice will melt completly.

So, let the final temperature be T

=(2.108*3*12)+(12*334)+12*4.18*T=420*4.18(24-T)

=4083.88+50.16T=1755.6(24-T)

=38050.52=1805.76T

=21.07 ⁰C