Question

A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting on a weight of mass 32 kg. The weight falls slowly through a distance of 5 m in driving the stirrer. Assume that all work done on the weight is transferred to the water and that the local acceleration of gravity is 9.8 m·s−2, determine:
(a) The amount of work done on the water.
(b) The internal-energy change of the water.
(c) The final temperature of the water, for which Cp =4.18 kJ/kgC.
(d) The amount of heat that must be removed from the water to return it to it initial temperature.

Answer 1

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the  work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The  change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.