Question

Some ice at -15 degrees celcius with a mass of 0.45 kg is
dropped into a highly insulated cup (no heat flows into or out of
cup) with 3.0 kg water that is at 20 degrees celcius.

Determine if all the ice has melted

What if the final temperature of the water in the cup?

Answer #1

Here, total heat energy is conserved. So, for melting of all ice, H= 0.45*2.1*(0-(-15)) { where 2.1 kJ/kgK is the specific heat capacity of ice} = 14.175kJ heat is to be provided by the colling of water.

Now, heat released by cooling of water to 0 degrees Celcius is H1= 3*4.18*(20-0) { where where 4.18 kJ/kgK is the specific heat capacity of water } = 250.8kJ. So, the heat released by cooling of water is more i.e. H1>H. So the ice will melt.

Now, for final temperature, the total heat exchange is same i.e. H= H1.

So, 0.45*2.1*(T-(-15))= 3*4.18*(20-T) { where T is the final temperature of the water in the cup, because after the ice melts it becomes water and both attain a common temperature }

So, T+15= 13.27 * (20-T)

or, T+15= 265.4- 13.27T

So, 14.27T= 250.4

So, the final temperature of the water in the cup is T= 17.54 degrees celcius.

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