Question

A 8.6-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.1 kg of water (cwater = 4186 J/kg-K) with an initial temperature of 293 K. 1) What is the final temperature of the water-and-cube system? K Submit 2) If the temperature of the copper was instead 1350 K, it would cause the water to boil. How much liquid water (latent heat of vaporization = 2.26 × 106 J/kg) will be left after the water stops boiling? kg Submit 3) Let's try this again, but this time add just the minimum amount of water needed to lower the temperature of the copper to 373 K. In other words, we start with the cube of copper at 750 K and we only add enough water at 293 K so that it completely evaporates by the time the copper reaches 373 K. Assume the resulting water vapor remaining at 373 K. How much water do we need ?

Answer 1

Given

Mc = 8.6 kg ; Cc = 386 J/kg -K ; Tc = 750 K

Mw = 5.1 kg ; Cw = 4186 J/kg -k ; Tw = 293 K

1)Let Tf be the final temperature of the system

- Q(lost) = Q(gained) (negative sign indicates heat is lost by Cu)

- Mc Cc (Tf - Tc) = Mw Cw (Tf - Tw)

- 8.6 x 386 (Tf - 750) = 5.1 x 4186 x (Tf - 293)

- 3319.6 Tf + 2489700 = 21348.6 Tf - 6255139.8

Tf = 354.51 K

Hence, Tf = 354.51 K

2)we know that the water boils at 100 deg C = 273 + 100 = 373 K

amount of heat to raise the temp of water from 293 to 373 will be

Qw = Mw Cw (T2 - T1) = 5.1 x 4186 x (373 - 293) = 1707888 J

Heat required to lower down the Cu cube from 1350 to 750 K

Qc = Mc Cc (T2 - T1) = 8.6 x 386 x (1350 - 373) = 3243249.2 J

Q(vap) =  3243249.2 J -  1707888 J = 1535361.2 J

we know latent heat of vaporization of water is :

L = 2.26 x 10⁶ J/kg

The mass of water vaporized is :

Mv =  1535361.2 J / 2.26 x 10⁶ J/kg = 0.68 kg

Mass of left over water will be:

M = Mw - Mv = 5.1 - 0.68 = 4.42 kg

Hence, M = 4.42 kg.

3)

Qc = Mc Cc (Tc - 373) = 8.6 x 386 x (750 - 373) = 1251489.2 J

Amount of heat required to raise the temp of M kg of water from 293 to 373 will be:

Qw = M Cw (373 - Tw) = 1 x 4186 x (373-293) = 334880M J

we know latent heat of vaporization of water is :

L = 2.26 x 10⁶ J/kg

Qc - Qw - MxL = 0

Qw + ML = Qc

334880M J + 2260000 M J = 1251489.2 J

M =  0.48 kg

Hence, mass of water needed = M = 0.48 kgs.