Answer to Question #157121 in Molecular Physics | Thermodynamics for Eva

Question

A well lagged aluminium calorimeter of mass 80g contains 150g of water and 100g of ice at 0^oC. A heating coil rated 1.0kw is put in the calorimeter and the mixture is stirred until its temperature is 33^oC. Calculate how much ice is left after 1 minute. State any assumption you made.

Answer 1

Mass of aluminum, $m_a = 80g = 0.08Kg$

Mass of water, $m_w = 150 g = 0.15 Kg$

Mass of ice, $m_i = 100g = 0.1 Kg$

Specific heat of water, $s_w = 4182K/Kg/K$

Specific heat of aluminum, $s_a = 0.9 J/g K = 900J/Kg/K$

Specific heat of ice, $s_i = 2.04KJ/Kg/K = 2040J/Kg/K$

Latent heat of ice, $L_f = 3.33 \times 10^5 J/Kg$

Initial temperature, $T_i = 0^\circ C$

Final temperature, $T_f = 33^\circ C$

Heat generated by the coil in one minute, $Q = 1000 \times 60 J = 60000 J$

Let x amount of the ice that melt,

then assume that there is no loss of the heat,

$(m_a s_a+m_ws_w+xs_i)(T_f-T_i)+m_iL_f =60000$

Putting values,

$[(0.080 \times 900)+(0.15 \times 4182)+(x \times 2040)](33-0) +(0.1 \times 3.33 \times 10^5) = 60000$

Solving it, we get, $x =0.05 Kg = 50 g$

Amount of ice left, $M = 100-50 = 50 g$