A 1.0 kg lead bar (c = 0.13 kJ/kg.K) at 80degreeC is placed in 2.0 kg of water (
ID: 1517533 • Letter: A
Question
A 1.0 kg lead bar (c = 0.13 kJ/kg.K) at 80degreeC is placed in 2.0 kg of water (c = 4.19 kJ/kgK) at 20degreeC. The final temperature of the lead bar is 21degreeC 28degreeC 41degreeC 50degreeC The amount of heat required to convert 1 g of ice and 0degreeC to steam at 100degreeC is C_ice=-0.5 cal/g-Cdegree, Lr(water_- 80 cal/g C_water = 1 calg-Cdegree), L, (water = 540 cal/g) 80 cal 180 cal 540 cal 720 cal A 2-mole ideal gas system is maintained at constant volume of 4 liters. If 200 J of heat is added, what is the increase in internal energy of the system? Zero 50 J 100 J 200 JExplanation / Answer
1)
Temperature of lead = 80 + 273 = 353 K
Temperature of water = 20 + 273 = 293 K
let the final temperature be T
therefore
heat lost by lead = heat gained by water
mass of lead x c of lead x (353 - T) = mass of water x c of water x (T - 293)
2 x 130 x (353 - T) = 2 x 4190 x (T - 293)
=> T = 294.8 k = 21.8C
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