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3 years ago

General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.

Chemistry

1 answer:

Helen [10]3 years ago

3 0

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

Q: heat absorbed

C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)

ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

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You have 750 grams of water at 80° Celsius. Which of the following would lower the temperature of the water by 10° Celsius? (1 p

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Answer:

adding 750 grams of water at 60° Celsius .

Explanation:

We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat lost by water (Q = ??? J).

m is the mass of water (m = 750.0 g).

c is the specific heat capacity of the water (c = 4.18 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.

Now, we can calculate the Q that is gained by the different added amounts of water:

adding 750 grams of water at 50° Celsius :

ΔT = 70.0°C - 50.0°C = 20.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.

adding 325 grams of water at 60° Celsius :

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.

adding 750 grams of water at 60° Celsius :

ΔT = 70.0°C - 60.0°C = 10.0°C,

∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.

adding 1000 grams of water at 55° Celsius:

ΔT = 70.0°C - 55.0°C = 15.0°C,

∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.

So, the right choice is:

adding 750 grams of water at 60° Celsius

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