CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J kg−1 K−1 and 4200 J kg−1 K−1 respectively. Neglect any thermal expansion. (a) How much heat is transferred to the liquid-vessel system? (b) How much work has been done on this system? (c) How much is the increase in internal energy of the system?

Open in App
Solution

Given:
The system comprises of an insulated copper vessel that contains water.
t1 = 15°C, t2 = 17°C

t1 is the initial temperature of the system
t2 is the final temperature of the system

∆t = Change in the temperature of the system = t2 − t1
= 17°C − 15°C = 2°C = 275 K

Mass of the vessel, mv = 100 g = 0.1 kg
Mass of water, mw = 200 g = 0.2 kg
Specific heat capacity of copper, cu = 420 J/kg-K
Specific heat capacity of water, cw = 4200 J/kg-K

(a) Since the system is insulated from the surroundings, no heat is transferred between the system and the surroundings. This implies that the heat transferred to the liquid vessel system is zero. The internal heat is shared between the vessel and water.

(b) Work done on the system=mwcwt+mvcut
⇒ dW = 100 × 10−3 × 420 × 2 + 200 × 10−3 × 4200 × 2
⇒ dW = 84 + 84 × 20 = 84 × 21
⇒ dW = 1764 J

(c) Using the first law of thermodynamics, we get

dQ=dW+dUHere, dW=pdV

Work is done by the system. Thus, work done is negative.
⇒ dQ = 0 (given)
dU = − dW
= -(-1764) = 1764 J

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
First Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon