Question

Suppose you have 600.0 grams of room temperature water (20.0 degrees Celsius) in a thermos. You drop 90.0 grams of ice at 0.00 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)?

Answer 1

Answer:

(a) what is the equilibrium temperature of the system?

since, heat is the energy that flows from the high temperature to the low temperature. So the heat which was lost by the water is grained by the ice.

so heat which has lost=mcΔT

                                    =600gm*1cal/gm.C*$(20^{0} C-0^{0} C)$

                                    =12000cal

Again for the ice to melt the heat gained by the ice is=m*L(f)

                                                                                        =90gm*80Cal/gm

                                                                                        =7200Cal

so the equilibrium temperature will be $0^{0} C$

(b) how much ice is left(in grams)?

Since, we know that the heat which the water has lost will be equal to the mass of ice*latent heat of freezing of the ice

12000cal=m1*80Cal/gm

m1=150gm.

so the mass of the ice which remained is=150-150=0gm.

Answer 2

Answer:

6.956 degree and 0 gm ice left

Explanation:

a) Heat flow from hot body to cold body. So here heat loss by water that is used to melt the ice.

Thus, heat lost by water= mcΔt

C= specific heat that is 1 cal / gram / k

Total heat water can lose to ice= 600 × 1 × 20.

= 12000 cal

The heat gained by the ice = mass × Latent heat

Latent heat of fusion= energy required for 1gm to change the state without changing temp = 80 cal / gm

Heat required by 90g ice to melt= 90 × 80

= 7200Cal

Heat left 12000-7200=4800 cal.

These 4800 calories raise the temp of 690 gm water (600 + 90)

Heat = mct

4800= 690×1×t

t=4800/690=6.956 degree