SFNUser Posted October 2, 2008 Posted October 2, 2008 If your hot coffee loses 50 kJ of energy as it cools and the volume shrinks because of thermal contraction, how large is the change in the internal energy of the coffee? How large is the change in the enthalpy? Explain. a) +50 kJ b) -50 kJ c) >50 kJ d)>-50kJ e)<50 kJ f)<-50 kJ (I really don't know where to get started on this problem. If someone could give me a thorough explanation, I'd greatly appreciate it!) Thanks in advance!
hermanntrude Posted October 7, 2008 Posted October 7, 2008 you need to know a few things first: the first law of thermodynamics: [math]\Delta U = q + w[/math] where [math]\Delta U[/math] is the change in internal energy and w is the work done BY the system and q is the heat absorbed by the system. The conventions for the signs are very important. When the system (the coffee cup, in this case) absorbs heat, the sign of q is positive. If the system gives off heat (loses it) the sign of q is negative. when the system does work on its surroundings (expands), the sign of w is negative, when a system has work done on it (it contracts), the sign of w is positive that should give you enough to figure it out
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