Question

As illustrated beside there is a glass beaker inside a thermally insulated box. In addition, the system contains an impeller driven by the pull of an external weight and designed to dissipate mechanical energy. Before closing the box you pour 1 L of water at 20 ◦C and then add 100 g of ice at -8 ◦C in the beaker.

• A) Once the box is sealed, what happens inside the beaker? If you wait a few hours, what is the composition inside the beaker? Is there only water, only ice or still a mix of ice and water? If your answer includes the presence of ice then calculate how much ice remains. In any case, calculate the temperature of whatever is contained in the beaker

Assume that, after waiting for a few hours, you decide to bring the content of the beaker to the original temperature of the water, that is 20 ◦C.

• B) Since the beaker remains inside the well insulated box, are you even able to increase its temperature? Can you do it by driving the impeller? If you answer yes then calculate how much work you need to provide to the impeller to bring the whole content of the beaker up to 20 ◦C.

• C) Assume that you want to do such work by attaching a weight, as illustrated beside, to drive the impeller. But you decide that you want to do it in a single stroke and the maximum height you can let your weight fall is 1.4 m. In this case, what should be the mass (in tonnes) of such weight?

Answer 1

m_i = mass of ice = 100 g

M = mass of water = 1000 g

c_i = specific heat of ice = 2.108

c_w = specific heat of water = 4.186

T_ii = initial temperature of ice = - 8

T_wi = initial temperature of water = 20

T = final equilibrium temperature

L = latent heat of fusion of ice = 334

using conservation of heat

heat gained by ice = heat lost by water

m_i c_i (0 - T_ii ) + m_i L + m_i c_w (T - 0 ) = M c_w (20 - T )

(100 x 2.108) (0 - (- 8)) + (100 x 334) + (100 x 4.186)(T - 0 ) = (1000 x 4.186) (20 - T )

T = 10.6 C

there is only water at temperature of 10.6 C

b)

W = work done

m = mass of water = 100 + 1000 = 1100 g

c = specific heat = 4.186

$\Delta$ T = change n temperature = 20 - 10.6 = 9.4 C

work done is same as the heat required to raise the temperature of water

hence

W = Q

W = m c $\Delta$ T

W = (1100) (4.186) (9.4)

W = 43283.24 J

C)

m = mass of the weight

W = potential energy of the weight

W = mgh

43283.24 = m (9.8 x 1.4)

m = 3154.76 kg

m = 3.15 tonnes