224.3 g of water (c = 4189 J/kg•K) is contained in a thermally insulated 92.0 g

Question

224.3 g of water (c = 4189 J/kg•K) is contained in a thermally insulated 92.0 g copper (c = 387 J/kg•K) bowl. Both the water and bowl are at 11.8 °C. A 85.8 g aluminum (c = 900 J/kg•K) ball at 72 K is dropped into the water.

1)After coming to equilibrium, what will the final temperature of the system be (in °C)?

2)How much water will be turned into steam (in g)?

3)How much water will be turned into ice (in g)?

4)How much water remains (in g)?

Please answer all the questions or don't answer them at all. Thank you

Explanation / Answer

Given

mass of

Water = 224.3 g = 0.224 kg,
copper = 0.092 kg,

aluminium = 0.0858 kg

specific heat of

water cW = 4189 J/kgk

copper Cc = 387 J/kgk

aluminium Ca = 900 J/kgk

temperature of

water and copper = 11.8 0C = 284.8k

aluminium = 72 k

here heat energy lost by water and copper = heat gain by aluminium

(T-T1)(mw*Cw+mc*Cc) = ma*Ca(T-T1)

(T -284.8)(0.224*4189+0.092*387)= 0.0858*900(72-T)

T =235.85
T = 273.15-235.85

T = 37.3 0C

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