A 1.0 kg lead bar (c=0.13 kJ/kg K) at 80 degree C is placed in 2.0 kg of water (
ID: 1517552 • Letter: A
Question
A 1.0 kg lead bar (c=0.13 kJ/kg K) at 80 degree C is placed in 2.0 kg of water (c= 4.19 kJ/kg K) at 20 degree C. The final temperature of the lead bar is 21 degree C 28 degree C 41 degree C 50 degree C The amount of beat required to convert 1 g of ice at 0 degree C to stream at 100 degree C is (C_ice =-0.5 cal/g-C^10, L_f (water)= 80 cal/g. C_water = 1 cal/g-C^12), L, (water = 540 cal/g)/80 cal 180 cal 540 cal 720 cal. A 2-mole ideal gas system is maintained at a 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) Here, Applying heat Law
=> ml * sl * (T - 80) + mw * sw * (T - 20) = 0
=> 1 * 130 * (T - 80) + 2 * 4190 * (T - 20) = 0
=> 8510T = 178000
=> T = 21 degree celsius ------------------> Final temperature
=> option a) is correct .
2) amount of heat required = 1 * 80 + 1 * 1 * 100 + 1 * 540
= 720 cal
=> option d) is correct
3) Here, increase in internal energy = Heat added to system
=> increase in internal energy = 200 J
=> option d) is correct .
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