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a 5 g ice cube starts life at -1 C if 2182.5 j are added to it what will be the final temperature upon his demise ?

Sagot :

Drpelezoidn

Answer:

Final Temp = 23.92°C

Explanation:

ΔH(total) = (m·c·ΔT)ice + (m·ΔH(f))melt'g + (m·c·ΔT)water

2182.5j = (5g)(2.092j/g·°C)(1°C) + (5g)(334.56j/g) + (5g)(4.184j/g·°C)(ΔT)

(2182.5 - 10.46 - 1672.8)j = 20.92j/°C·ΔT

ΔT = (2182.5 - 10.46 - 1672.8)j / 20.92j/°C = 23.92°C

Since the melting ice starts and ends at 0°C and is then warmed to 23.92°C then the temperature change is also the final temp of the water based upon given energy input values.

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