Question

1.45 kg    of ice at   -5 ∘C    is dropped into   31 kg    of water at temperature   15.3 ∘C   .

Part A - Final Temperature

What is the final temperature of the system? (If I've set the problem up right, you should be given numbers such that all of the ice has melted.)

Tf   =		∘C

Part B - Entropy of Warming Ice

What is the change in entropy of the ice, as it warms up before melting?

S =		J/K

Part C - Entropy of Ice Melting

What is the change in entropy of the ice as it melts?

S =		J/K

Part D - Entropy of Water From Ice Warming Up

After the ice melts, it's just water, and this water still has to warm up to the final temperature. What is the entropy change of this process?

S =		J/K

Part E - Change in Entropy of the Water Cooling

Meanwhile, the initially warm water is cooling down. What is the entropy change of this water, which is undergoing cooling?

S =		J/K

Part F

What is the total change in the entropy of the system?

S =		J/K

Answer 1

mass of ice m = 1.45 kg

Initial temparture of ice t = -5 ∘C

mass of water M= 31 kg

temperature of water t ' = 15.3 ∘C   .

Heat lost by water = heat gain by ice

MC(t ' - T ) = mc(t-0) + mL+mC(T-0)

                = m[ct+L+cT]

Where C = Specific heat of water = 4186 J / kg oC

          c = Specific heat of ice = 2100 J / kg oC

          L = latent heat of fusion of ice = 333.4 x10 ³ J / kg

Substitute values you get

31 x4186 (15.3-T) = 1.45[(2100x5)+(333.4x10 ³)+4186 T]

(31x4186 /1.45)(15.3-T) = [(2100x5)+(333.4x10 ³)+4186 T]

89493.8(15.3 -T) = 343900 +4186 T

1.3692 x10 ⁶ -89493.8 T = 343900 +4186 T

   1025300 = 93679.8 T

            T = 10.95 ^oC

(b). Change in entropy dS = mc ln(T ' / t ) where T ' = 0 oC = 273 K ,t = -5 oC = -5+273 = 268 K

                                      = 1.45x2100x ln (273 /268)

                                      = 56.28 J/K

(c).Change in entropy dS ' = mL / T '

                                       = (1.45 x333.4 x10 ³ ) / 273

                                       = 1770.8 J/K

(d).Change in entropy dS" = mC ln( T / T ')     where T = 10.95 oC = 10.95 + 273 = 283.95 K

                                       = 1.45 x4186 x ln( 283.95 /273)

                                       = 238.7 J/K

(e).Change in entropy ds = MC ln( T / t ' )   where t' = 15.3 o C = 15.3 + 273 = 288.3 K

                                     = 31 x4186 xln(283.95 / 288.3)

                                     = -1972.9 J/K

(f).Total change in entropy = dS + dS ' +dS" + ds