Exactly 15.2 mL of water at 26.0^degree C are added to hot iron skillet. All of

Question

Exactly 15.2 mL of water at 26.0^degree C are added to hot iron skillet. All of the water is converted into steam at 100^degree C. The mass of the pan is 1.20 kg and the molar heat capacity of iron is 25.19 J/(mol middot C). What is the temperature change of the skillet? Number -83.89^degree C You actually need to know the initial temperature of the iron skillet since you are only asked for Delta T, not the actual final T. All you you need to do is calculate the total amount of heat needed to BOTH raise the temperature of the water to 100^degree C and then boil it. This total amount of heat came FROM the skillet and must cause the temperature of the skillet to decrease.

Explanation / Answer

V = 15.2 mL water = 15.2 g

Tw = 26°C

all water goes to steam at 100°C

mass of pan = 1.20 kg = 1200 g of iron

Cp = 25.19 J/gC

Find dT of skillet

Qwater = Qliquid + Qvapor = m*Cp*(Tf-Ti) + m*Latent Heat of vaporization

Qwater = 15.2*4.184*(100-26) + 15.2*(2264.76 ) = 39130.5152 J

total Q transferred --< 39130.5152 J

Qmetal = m*C*(Tf-Ti)

39130.5152 = 1200*25.19*dT

dT = 39130.5152/( 1200*25.19) = 1.2945°C

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