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den301095 [7]
3 years ago
11

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

esulting combination will have a temperature of 62.5°C? Assume that coffee and water have the same density and the same specific heat (4.18 J/g·°C) across the temperature range.
Physics
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}

The heat (Q) can be calculated using the following expression:

Q=c \times m \times \Delta T

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

162mL.\frac{0.997g}{mL} = 162g

Then,

Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g

The volume of water is:

139g.\frac{1mL}{0.997g} =139mL

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Answer:

Incomplete question:

Refer to the temperature versus time graph when answering the questions in Parts C through F. A system consists of 250 of water. The system, originally at = 21.0 , is placed in a freezer, where energy is removed from it in the form of heat at a constant rate. The figure shows how the temperature of the system takes to drop to, after which the water freezes. Once the freezing is complete, the temperature of the resulting ice continues to drop, reaching temperature after an hour. The following specific heat and latent heat values for water may be helpful.

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Explanation:

To solve this exercise it is necessary to know that if the system is a single phase in which there is a temperature change or if it is a phase change at a single temperature. In the first case, the following formula would be used to calculate the amount of heat:

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Here

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Substituting:

Q₂ = 250*333.7 = 83425 J

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Qtotal = 154350+83425=237775 J

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