Question

Name Date First Law of Thermodynamics Calculations: Apply the First Law of Thermodynas following problems. 1. 2260 joules of heat is added to a svatem and 1500 2 joules of work is done by the system. What is the change in internal energy of the system? 3500 J of heat is added to a system and 500 is done by the system. What is the change in it energy of the system? 3. 4200 joules of work is done on a system and 50004. A system does 2500 J of work. If the change in joules of heat is lost. What is the change in internal internal energy of the system was 3520 J, how much energy of the system? heat did the system lose/gain?

Answer 1

law of thermodynamice, According to first All = Q-w where AU= change in internal energy Q = heat added W = work done by the system. 1) 2250 J of heat 1500 J of work is added is done to a system. by the system. Change in internal, 10 = Heat added (0) - work done by the system (w) energy su = 2250 J - 15005 (su = 7505 2) 3500 J of heat is added 5005 of work is done by to a system. the system. Change in internal energy, sue = heat added cm) - work done by the syetem (W) AU = 3500J - 5005 [AU = 3000

system 4200) of work is done on a 5000J of heat is lost » Heat is lost, so Q = - 5000] y work is done on the system, so w = -42005 A AU=Q-w su = -5000 - (-4200) su = -5000 + 4200 Du = -800) QU = 8005] → "-ve" sign indicates that 8005 of energy is lost from the system su=8005 4) A System does 25005 o work. ie, 25005 of work is done by the system. change in internal energy, su = 35205 su=Q-W 3520 - Q -2500 Q - 3520 +2500 Q = 60205) 60205 of heat is lost or gained by the system.