Question

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 31.2◦C. In an attempt to cool the liquid, which has a mass of 196 g, 131 g of ice at 0.0◦C is added. At the time at which the temperature of the tea is 26.5◦ C , find the mass of the remaining ice in the jar. The specific heat of water is 4186 J/kg ·◦ C . Assume the specific heat capacity of the tea to be that of pure liquid water. Answer in units of g.

Answer #1

Let x be the mass of the ice that already melted. The thermal energy of the tea at 31.2 °C must equal the thermal energy of the tea plus the water from the ice at 26.5°C plus the energy required to melt x.

"When ice melts, it absorbs as much heat energy (the heat of fusion) as it would take to heat an equivalent mass of water by 80 °C, while its temperature remains a constant 0 °C."

Therefore:

4186 * 0.196 * 31.2 = 4186 * (0.196 + x) * 26.5 + 4186 * x * 80

0.196 * 31.2 = 0.196 *26.5 + x * ( 26.5 + 80)

x = 0.196 * (31.2 - 26.5) / (26.5 + 80) = 0.0086497 kg = 8.65 g

hope this helps.

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The remaining ice is 131 - x = 122.35 g...answer

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