33 . Judy places 0.150 kg of boiling water in a thermos bottle. How many kilogra
ID: 1409241 • Letter: 3
Question
33.Judy places 0.150 kg of boiling water in a thermos bottle. How many kilograms of ice at –12.0 °C must Judy add to the thermos so that the equilibrium temperature of the water is 75.0 °C? 33.Judy places 0.150 kg of boiling water in a thermos bottle. How many kilograms of ice at –12.0 °C must Judy add to the thermos so that the equilibrium temperature of the water is 75.0 °C? 33.Judy places 0.150 kg of boiling water in a thermos bottle. How many kilograms of ice at –12.0 °C must Judy add to the thermos so that the equilibrium temperature of the water is 75.0 °C?Explanation / Answer
Principle of calorimetry
heat given by water = heat recieved bt ice
mw * cw * (Tw - T) = mi * ci * (T0 - Ti) + mi * L + mi * cw * (T - T0)
mass of boiling water mw = 0.150 kg
specific heat of water cw = 4.2 * 103 J/kg-0C
Initial temperature of boiling water Tw = 100 0C
mass of ice = mi
specific heat of ice ci = 2.05 * 103 J/kg-0C
Latent heat of melting of ice L = 334 * 103 J/kg
Initial temperature of ice Ti = - 12 0C
Melting point of ice T0 = 0 0C
Equilibrium temperature T = 75 0C
0.150 * 4.2 * 103 * ( 100 - 75)
=mi * { 2.05 * 103 * (0 - ( - 12)) + 334 * 103 + 4.2 * 103 * (75 - 0)}
mi = 15.75 / 673.6
= 0.0233 kg
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