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What is the amount of heat (in Joules) absorbed by 18 g of water initially at room temperature heated to 100°C? If 10 g of Cu is added to this water, than decrease in temperature (in Kelvin) of water was found to be?

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What is the amount of heat (in Joules) absorbed by 18 g of water initially at room temperature heated to 100°C? If 10 g of Cu is added to this water, than decrease in temperature (in Kelvin) of water was found to be? C (p,m) for water 75.32 J/mol K ; C (p,m) for Cu = 24.47 J/mol K.

(a) 5649, 369

(b) 5544, 324

(c) 5278, 342

(d) 3425, 425

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Correct answer: (a) 5649, 369

Explanation:

18gm of water at 100°C

10gm of Cu at 25°C is added

qp = Cp,mdT

= 75.32 x \(\frac{J}{K\;mol}\) x \(\frac{18g}{18g/mol}\)(373 - 298)K

= 75.32\(\frac{J}{K}\) x 75 K

= 5.649 x 103J

If now 10g of copper is added

Cp, m = 24.47 J/ mol K

Amount of heat gained by Cu

= 24.47\(\frac{J}{K\;mol}\) x \(\frac{10g}{63g/mol}\)(373 - 298)K

= 291.3 J

Heat lost by water = 291.30 J

-291.30 J = 75.32 \(\frac{J}{K}\) x \((T_2 - 373K)\)

⇒ -3.947 K = T2 - 373 K

⇒ T2 = 369.05 K

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