1st Law of Thermodynamics

Q=50∘

Question

Find the value of \(\Delta U\).

\( K_{Q=50 J } \quad 30 J \)

+20 J

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-80 J

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+80 J

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-20 J

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Solution

The correct option is B -80 J
Take a minute and try to get this on your own. Then move on the next video for the explanation.

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Similar questions

P = 50 \sum r = 1 \frac{^{50 + r} C_{r} (2 r - 1)}{^{50} C_{r} (50 + r)}, Q = 50 \sum r = 0 (^{50} C_{r})^{2}

R = 100 \sum r = 0 (- 1)^{r} (^{100} C_{r})^{2}

A B | | C D, ∠ A P Q = 50^{\circ} a n d ∠ P R D = 127^{\circ}

p = \sqrt{32} - \sqrt{24}

q = \sqrt{50} - \sqrt{48}

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