A worker pours 1.210 k g of molten lead at a temperature of 327.3 ? C into 0.498

Question

A worker pours 1.210kg of molten lead at a temperature of 327.3 ?C into 0.4980kg of water at a temperature of 75.00 ?C in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium. Express your answer using four significant figures. A worker pours 1.210kg of molten lead at a temperature of 327.3 ?C into 0.4980kg of water at a temperature of 75.00 ?C in an insulated bucket of negligible mass. Assuming no heat loss to the surroundings, calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium. Express your answer using four significant figures.

Explanation / Answer

Assuming water begins to boil and a part of it evaporates

M(lead)*c(lead)*(327.3-100) =M(water)*c(water)*(100-75) + Mx*lambda

c(lead) =0.16 J/g/K

c(water) =4.18 J/g/K

lambda =2260 J/g for water

1210*0.16*227.3 =498*4.18*25 +Mx*2260

44005=52041+Mx*2260

Mx is negative it means the water does not boil. The masses are the same as initial

M(water) =0.4980 kg

M(lead) =1.2100 kg

Computing final temperature

1210*0.16*(327.3-T) =498*4.18*(T-75)

63365.28-193.6*T =2081.64*T-156123

T =96.4682 degree Celsius

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