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Suppose you have 600.0g of room temperature water (20.0C) in a thermos. You drop 90.0gof ice at 0.00C into the thermos and shut the lid.
(a) What is the equilibrium temperature of the system?
(b) How much ice is left (in grams)?

Answer
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Hint: As we know that heat is the energy that flows from the high temperature to low temperature, so the heat lost by water will be gained by the ice which will be equal to the latent heat of freezing of ice.

Complete Step by step answer:As we know that heat is the energy that flows from the high temperature to low temperature, so the heat lost by water will be gained by the ice which will be equal to the latent heat of freezing of ice.
We are given with mass of water that is 600.0g and the temperature is (20.0C)and we first need to calculate the amount of heat lost and for that we know that heat capacity is given by the formula:
Q=mcΔT where m is the mass, c is calories and ΔT is change in temperature.
After we put the values in the formula, we will get:
Q=600.0×1×(200C)
Q=12000cal
And again for the ice to melt the heat gained by the ice is equal to mL(f)where L is the latent heat so we get,
Q=90.0×1×(100C20C)Q=7200cal
Therefore, the equilibrium temperature will be 0C.
(b)Now, since we know that the heat which the water has lost will be equal to the mass of the ice and latent heat of freezing of the ice, so we can calculate the mass as:
12000cal=m×1×80calgm1
m=150g

So the mass of the ice which remained is 150150=0g

Note: Whenever the heat energy is supplied to the system usually the temperature increases except during phase transition and rise in temperature is different for different substances and depends on heat capacity. Heat capacity is an extensive property since it depends on the amount of the matter present.