Question

Three hundred grams of ice at 0 degree C is added to 200 grams of water at 95 degree C. The system is kept thermally insulated from its environment. (a) How many joules of energy will it take to melt the ice? (b) What is the equilibrium temperature if the ice in (a) is placed in the water in part (b)? What phases are present? Ten grams of molten lead at 480.0 degree C are added to a water bath originally at 0.00 degree C. The system is kept thermally insulated from its environment so no energy is lost. (c) What is the minimum amount of water needed to cool the lead to a temperature of 99.0 degree C?

Answer 1

Solution

a) Latent heat of fusion of ice, H_f= 334 J/g

Energy tried to melt the ice, E= m*H_f= 300*334 J = 100200 J

b) using, Q= m*c* $\Delta T$

Let T be the equilibrium temperature.

Hear loss= Heat Gain

200*4.18*(95-T) = 100200 + 300*4.18*(T-0)

T can not be negative

So, T is 0 °C

heat loss by hot water, Qh= 200*4.18*(95-0) = 79800 J

So, ice will not completely melt.

phase present are solid and liquid.

c) Melting point of lead is 327 °C

specific heat of lead, c' = 0.128 J/(g°C)

Latent heat of fusion of lead, H= 23 J/g

Let m be the mass of water needed.

heat loss=heat gain

10*0.128*(480-327) + 10*23 + 10*0.128*(327-99) =m*4.18*(99-0)

or, mass of water required, m= 1.8 g