A 7.1-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dro

Question

A 7.1-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.9 kg of water (cwater = 4186 J/kg-K) with an initial temperature of 293 K. 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

let the mass of water needed be m kg.

then heat reelased by copper=7.1*386*(750-373)=1033.2 kJ

now to make water heated from 293 K to 373 K, heat required=m*4186*(373-293)=334.88*m kJ

to evapaorate completely, heat required=m*2260 kJ

so 1033.2 =334.88*m+2260*m

m=0.398 kg=398 grams

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