Physics, asked by gk863024, 11 months ago

Suppose you have 500.0 grams of room temperature water (20 degrees Celsius) in a thermos. You drop 150.0 grams of ice at 0 degrees Celsius into the thermos and shut the lid.
(a) What is the equilibrium temperature of the system?
(b) How much ice is left (in grams)?

Answers

Answered by AneesKakar
7

Answer:

0°C and 25 gram.

Explanation:

(a) Since, heat is the energy that flows from the higher temperature to the lower temperature. So, the heat which was lost by the water is gained by the ice.

Heat which water has lost = mc∆T, in which c is specific heat of water.

= 500gm × 1cal/gm.°C × (20°C - 0°C) = 10000 cal(since the water;s temperature has dropped to 0°C)

Again, for the ice to melt the heat gained by the ice is = m*L(f) , here the L(f) is latent heat of freezing of the water.

= 150gm × 80cal/gm

= 12000 cal.

Since, 1200>10000 so we get that the ice which did not melt completely concluding that the system temperature is 0°C.

Therefore the equilibrium temperature will be 0°C.

(b)Since, we know that the heat which the water has lost will be equal to the  mass of ice melted × latent heat of freezing of the ice.

10000 Cal = m1 × 80 cal/gm.

m1 = 10000/80 = 1000/8 = 125gm.

So, the mass of ice which remained is = 150 - 125 = 25gm.

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