Question #24201

At a party 0.50 kg tube ice at -10 degrees C is placed in 3.0 kg of iced tea at 20 degrees C . At what temperature will be the final mixture be?The tea can be considered as water.Assume a heat flow sorroundings and the container.

Heat gain = Heat loss

Heat gain = Heat loss

Expert's answer

The heat offusion of water ice is 334 kJ/kg.

Ice heat capacity is 2.11 kJ / (kg * K)

Water heat capacity (for temperature above zero) is 4.1813 kJ / (kg * K).

Let's assume the final mixture is totally liquid at temperature T above zero in Celsius scale. Then the heat balance equation is

(0.50 kg) * (2.11 kJ / (kg * K)) * (10 K) + (0.5 kg) * (334 kJ/kg) + (0.5 kg) *

(4.1813 kJ / (kg * K)) * (T K) = (3.0 kg) * (4.1813 kJ / (kg * K)) * ((20 - T)

K)

From here,

T = ( - (0.50 kg) * (2.11 kJ / (kg * K)) * (10 K) - (0.5 kg) * (334 kJ/kg) +

(3.0 kg) * (4.1813 kJ / (kg * K)) * (20 K) ) / ( (0.5 kg) * (4.1813 kJ / (kg *

K)) + (3.0 kg) * (4.1813 kJ / (kg * K)) ) = 5.011 K

Answer: about 5 K.

Ice heat capacity is 2.11 kJ / (kg * K)

Water heat capacity (for temperature above zero) is 4.1813 kJ / (kg * K).

Let's assume the final mixture is totally liquid at temperature T above zero in Celsius scale. Then the heat balance equation is

(0.50 kg) * (2.11 kJ / (kg * K)) * (10 K) + (0.5 kg) * (334 kJ/kg) + (0.5 kg) *

(4.1813 kJ / (kg * K)) * (T K) = (3.0 kg) * (4.1813 kJ / (kg * K)) * ((20 - T)

K)

From here,

T = ( - (0.50 kg) * (2.11 kJ / (kg * K)) * (10 K) - (0.5 kg) * (334 kJ/kg) +

(3.0 kg) * (4.1813 kJ / (kg * K)) * (20 K) ) / ( (0.5 kg) * (4.1813 kJ / (kg *

K)) + (3.0 kg) * (4.1813 kJ / (kg * K)) ) = 5.011 K

Answer: about 5 K.

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