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NEET-XII-Physics

26: Laws of Thermodynamics

with Solutions - page 3
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  • #1
    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? (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?
    Ans : 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`` ={m}_{w}{c}_{w}∆t+{m}_{v}{c}_{u}∆t``
    ⇒ 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+dU``
    `` \,\mathrm{\,Here\,},dW=pdV``
    Work is done by the system. Thus, work done is negative.
    ⇒ dQ = 0 (given)
    dU = - dW
    = `` -``(`` -``1764) = 1764 J
    Page No 62: (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`` ={m}_{w}{c}_{w}∆t+{m}_{v}{c}_{u}∆t``
    ⇒ 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+dU``
    `` \,\mathrm{\,Here\,},dW=pdV``
    Work is done by the system. Thus, work done is negative.
    ⇒ dQ = 0 (given)
    dU = - dW
    = `` -``(`` -``1764) = 1764 J
    Page No 62:
  • #1-a
    How much heat is transferred to the liquid-vessel system?
    Ans : 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.
  • #1-b
    How much work has been done on this system?
    Ans : Work done on the system`` ={m}_{w}{c}_{w}∆t+{m}_{v}{c}_{u}∆t``
    ⇒ dW = 100 × 10-3 × 420 × 2 + 200 × 10-3 × 4200 × 2
    ⇒ dW = 84 + 84 × 20 = 84 × 21
    ⇒ dW = 1764 J
  • #1-c
    How much is the increase in internal energy of the system?
    Ans : Using the first law of thermodynamics, we get
    `` dQ=dW+dU``
    `` \,\mathrm{\,Here\,},dW=pdV``
    Work is done by the system. Thus, work done is negative.
    ⇒ dQ = 0 (given)
    dU = - dW
    = `` -``(`` -``1764) = 1764 J
    Page No 62: