A 7.7-kg cube of copper (c Cu = 386 J/kg-K) has a temperature of 750 K. It is dr

Question

A 7.7-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?

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?

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 ?

Explanation / Answer

1) Let the final temperature be T,

heat gained by water = heat lost by copper

5.1*4186*(T-293) = 7.7*386*(750-T)

21348.6 T - 6255139.8 = 2229150 -  2972.2T

24320.8T = 8484289.8

T = 8484289.8/24320.8

= 348.85 K

2) Final temperature = 273+100 K =373K

Heat lost by copper = 7.7*386*(1350-373)

= 2903839.4 J

let m mass of water of water is left

5.1*4186*(373-293) + (5.1-m)*2.26*10^6 =2903839.4

(5.1-m)*2.26*10^6 = 2903839.4 - 5.1*4186*(373-293) = 1195951.4

5.1 - m = 1195951.4/[2.26*10^6 ] = 0.529 kg

m = 5.1-0.529 = 4.571 kg

3) let the mass of water we need be m

heat gained by water = heat lost by copper

m*4186*(373-293) + m*2.26*10^6   = 7.7*386*(750-373)

m*[4186*(373-293) + 2.26*10^6 ] = 1120519.4

m = 1120519.4/ [4186*(373-293) + 2.26*10^6 ]

= 0.432 kg

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