Question

A thermally insulated closed copper vessel contains water at ${15^ \circ }C$ . When the vessel is shaken vigorously for 15 minutes, the temperature rises to ${17^ \circ }C$. 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 $420\,Jk{g^{ - 1}}{K^{ - 1}}$ and $4200\,Jk{g^{ - 1}}{K^{ - 1}}$ respectively. Neglect any thermal expansion.
(a) How much heat is transferred to the liquid-vessel system?
(b) How much work has been done on the system?
(c) How much is the increase in the internal energy of the system?

Answer 1

In an insulated system there is no exchange of heat with the surrounding. The work done in shaking the vessel is converted into heat energy. So, if we find the total heat produced, we will get the total work done. Since the change in heat, $dQ$ is zero, by using the first law of thermodynamics we get the change in internal energy equal to the work done.

Complete step by step answer:
The initial temperature of water is given as
${\theta _i} = {15^ \circ }C$
After time, $t = 15\,\min$ the temperature becomes
${\theta _f} = {17^ \circ }C$
Mass of copper vessel,
${m_c} = 100g = 0.1kg$
Mass of water,
${m_w} = 200g = 0.2kg$
Specific heat capacity of copper,
${S_c} = 420\,Jk{g^{ - 1}}{K^{ - 1}}$
Specific heat capacity of water,
${S_w} = 4200\,Jk{g^{ - 1}}{K^{ - 1}}$
Thermal expansion is the expansion due to heat. When we shake the vessel we are doing work. So heat is generated due to thermal inertia. So the temperature will increase. So the copper will expand due to this increase in temperature. Here since thermal expansion is neglected the volume remains constant.

(a) In the first part of the question, we are asked to find the heat transferred to the liquid-vessel system. It is given the copper vessel is thermally insulated and closed. So, no heat will be transferred to the system from the surrounding. The increase in heat is only due to the work that we do in shaking.

There is an internal exchange of heat between water and vessel but no energy is transferred to the system externally.

(b) In the second part we need to find the total work. Since the work done due to shaking is converted into heat energy, we can write
Work done = heat produced
We know the heat is calculated as $Q = mS\Delta \theta$
Where $m$ is the mass, $S$ is the specific heat, and $\Delta \theta$ is the change in temperature.
In the case of water ${Q_w} = {m_w}{S_w}\Delta \theta$ and for copper vessel ${Q_c} = {m_c}{S_c}\Delta \theta$.
$\Rightarrow \Delta W = {m_c}{S_c}\Delta \theta + {m_w}{S_w}\Delta \theta$
$\Rightarrow \Delta W = \left( {{m_c}{S_c} + {m_w}{S_w}} \right)\Delta \theta$
$\Rightarrow \Delta W = \left( {0 \cdot 1 \times 420 + 0 \cdot 2 \times 4200} \right) \times \left( {17 - 15} \right)$
$\Rightarrow \Delta W=1764$

$\therefore$ Total work done = 1764J$

(c) Since it is a closed vessel, change in heat, $dQ = 0$
From first law of thermodynamics we have
$dQ = dU + dW$
Since $dQ = 0$
$\Rightarrow dU = - dW$
So, the increase in internal energy will be equal to the work done.
Work done on the system is taken as negative. Thus, we get,
$\Rightarrow dU = - \left( { - 1764\,J} \right)$
$\Rightarrow dU = 1764\,J$

$\therefore$ The increase in internal energy is $1764\,J$.

Note:
Remember that an insulated system is a system with no exchange of thermal energy with surroundings. Also, remember the sign conventions to be used while calculating work. The work done on the system by external agencies is taken as negative. Whereas, the work done by the system is taken as positive.