What is the change in internal energy (Delta E) when a system is heated with 35

Question

What is the change in internal energy (Delta E) when a system is heated with 35 J of energy while it does 15 J of work? +50 J -20 J +20 J +35 K -50 J Heat is best defined as a substance that increases the temperature and causes water to boil. a form of potential energy the total energy that a substance has. a form of work. energy transferred as the result of a temperature difference. Use the following information to determine the standard enthalpy change when 1 mol of PbO(s) is formed from lead metal and oxygen gas. PbO(s) + C(graphitc) rightarrow Pb(s) + CO(g) Delta H degree = 107 kJ 2C(graphite) + O_2(g) rightarrow 2CO(g) -436 kJ +436 kJ -218 kJ -329 kJ -115 kJ

Explanation / Answer

1) Using the IUPAC convention (work is done on the system by the surroundings, hence positive), we have,

E = q + w where q denotes heat and w denotes work. Hence,

E = (35 + 15) J = + 50 J

However, if we use the Clausius convention (work is done by the system and hence negative), we have,

E = q – w = (35 – 15) J = +20 J

I am not sure which form of the first law (E expression) you have in your text book. Please consult the form of E given in your text and your answer will be either (a) + 50 J or (c) +20 J (ans)

2) Heat is best defined as (e) energy transferred as a result of temperature difference.

3) PbO (s) + C (graphite) --------> Pb (s) + CO (g)              H0 = 107kJ

    2 C (graphite) + O2 (g) --------> 2 CO (g)                         H0 = -222 Kj

Invert the first equation and multiply by 2 (sign of H0 will change but do not multiply H0 by 2) and add to the second equation. We will get

2 Pb (s) + O2 (g) ----------> 2 PbO (s)

H0 = -H0 (first equation) + H0 (second equation) = -107 kJ + (-222 kJ) = -329 kJ (ans)

Ans: (d) -329 kJ

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