Question

Two 20.0-g ice cubes at –13.0 °C are placed into 275 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, Tf, of the water after all the ice melts.

Heat capactiy of H2o (s) = 37.7 J/(mol x K)

Heat capactiy of H2o (l) = 75.3

enthapy infusion of H20= 6.01 kj/mol

Answer 1

Specific heat capacity of ice, Ci = 37.7 J/moloC

Latent heat of fusion of ice, Lf = 6.01 J/mol

Specific heat capacity of water, Cw = 75.3 J/moloC

mol of ice, mi = mass / molar mass = 20.0 g / 18 g/mol = 1.11 mol

mol of water, mw = 275 g / 18 g/mol = 15.28 mol

heat required to take ice from -13.0 oC to 0 oC,

Q1 = ni*Ci*Δ Ti

= 1.11*37.7*(0-(-13.0))

= 544 J

heat required to take water from 25.0 oC to 0 oC,

Q2 = nw*Cw*Δ Tw

= 15.28*75.3*(25.0-0)

= 28765 J

heat required to melt whole ice,

Q3 = ni*Lf

= 1.11*6.01

= 6.671 KJ

= 6671 J

Q1 + Q3 = 544+6671=7215<Q2

So whole of ice will melt

Only water will be there in the system

let final temperature be ToC

Q1 + Q3 + heat required by melted ice = heat released by water

544 + 6671 +ni*Cw*deltaT = nw*Cw*Δ Tw

544 + 6671 + 1.11*75.3*(T-0) = 15.28*75.3*(25.0-T)

7215 + 83.58*T = 28765 - 1150.6*T

1234.2*T = 21550

T = 17.5 oC

Answer: 17.5 oC