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
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