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?

Answer 1

(a) Heat transferred to the liquid-vessel system is :

• There is no transfer of heat to the liquid-vessel system as system is thermally insulated.

(b) Work has been done on this system is :

• Given, T1 = 15°C, T2 = 17°C, mass of vessel, Mv = 100g, mass of water, Mw = 200g

• The specific heat capacities of copper and water are 420 J kg−1 K−1 and 4200 J kg−1 K−1 respectively.

• ∆T = 17°C - 15°C = 2°C = 275K

• Work done on the system, ∆W = Mv×c(copper)×∆T + Mw×c(water)×∆T

= (0.1Kg)×(420 J kg−1 K−1)×275K + (0.2Kg)×(4200 J kg−1 K−1)×275K

= 1764 J

• ∆W = -ve ( work is done on the system )

∆W = -1764 J

(c) Increase in the internal energy of the system is :

• We know that ∆Q = 0

.•. ∆Q = ∆U + ∆W

0 = ∆U + ∆W

∆U = -∆W

= -(-1764 J)

= 1764 J

Answer 2

Explanation:

The system consists of a copper-insulated vessel containing water.

$\begin{aligned}&t_{1}=15^{\circ} \mathrm{C}\\&t_{2}=17^{\circ} \mathrm{C}\end{aligned}$

$\mathrm{M}_{\mathrm{V}}$ = 100 g = 0.1 kg

$\mathrm{M}_{\mathrm{W}}$= 200 g = 0.2 kg

$\begin{aligned}&c_{\mu}=420 \mathrm{J} / \mathrm{kg}-\mathrm{K}\\&c_{m}=4200 \mathrm{J} / \mathrm{kg}-\mathrm{K}\end{aligned}$

$t_{1}$= initial temperature  

$t_{2}=$ final temperature  

$\mathrm{M}_{\mathrm{V}}$ is Mass of the vessel,  

$\mathrm{M}_{\mathrm{W}}$ is Mass of water,  

$\mathrm{C}_{\mathrm{U}}$ is Specific heat capacity of copper

$\mathrm{C}_{\mathrm{w}}$ is Specific heat capacity of water

∆t   is Change in the temperature  

$\Delta t=t_{2}-t_{1}$

   = 17°C − 15°C = 2°C = 275 K  

 (a) since the network is isolated from the environment no heat is transmitted between the system and the environment. This means that the heat transferred to the system of liquid vessels is zero. The interior heat between the vessel and the water is shared.

(b) Work done  

$=m_{w} c_{w} \Delta t+m_{v} c_{u} \Delta t$

$=100 \times 10^{-3} \times 420 \times 2+200 \times 10^{-3} \times 4200 \times 2$

$=84+84 \times 20=84 \times 21$

$d W=1764 J$

(c) Using thermodynamics first law  

 $\begin{array}{l}{d Q=d W+d U} \\{d W=p d V}\end{array}$

System work done. So the work that is done is negative.

$\begin{array}{l}{\mathrm{dQ}=0} \\{\mathrm{dU}=-\mathrm{dW}} \\{=-(-1764)=1764 \mathrm{J}}\end{array}$