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I need help with Number 5, Thank you!
5. Suppose 150g of ice at a temperature of 0 C are combined with 250 g of water at 70°C.in tainer. The onty heat nges are bet
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Answer #1

a) For the system to cool down to 0 C, a certain amount of heat energy must be provided so that only a part of ice is melted and not the whole, otherwise the equilibrium will shift from 0 C

Let us calculate the energy required to melt whole of the ice

Q = m L where L is latent heat of fusion

= 150 gms × 80cal/gms =12000 cal

The energy supplied by water at max can be

Q = m c ∆T = 250×1×70 = 17,500 cal

So yes the water has sufficient energy, first to melt the ice and then raise the temperature of the system as a whole so final equilibrium temperature will not be 0°C

b) The final state of equilibrium will be purely water and no ice and the water in a beaker will be at a temperature greater than 0°C.

c) To calculate the final equilibrium temperature,

As we have seen, after converting whole of ice into water, 17500-12000 cal = 5500 cal will still be left.

So using heat equation.

Q = m s ∆T

5500 = 400 × 1 × ∆T ( m = 250 + 150= 400 as all ice has been converted into water)

∆T = 13.75 °C

Hence the equilibrium temperature of the system will be 13.75 °C with 400gm water and 0g ice.

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