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olga2289 [7]
3 years ago
15

The temperature of a 95.4 g piece of Cu increases from 25.0 °C to 48.0 °C when the Cu absorbs 849 J of heat. What is the specifc

heat of Cu​
Chemistry
1 answer:
melisa1 [442]3 years ago
3 0
<h3>Answer:</h3>

0.387 J/g°C

<h3>Explanation:</h3>
  • To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
  • Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
  • That is, Q = mcΔT

in our question we are given;

Mass of copper, m as 95.4 g

Initial temperature = 25 °C

Final temperature = 48 °C

Thus, change in temperature, ΔT = 23°C

Quantity of heat absorbed, Q as 849 J

We are required to calculate the specific heat capacity of copper

Rearranging the formula we get

c = Q ÷ mΔT

Therefore,

Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)

                                        = 0.3869 J/g°C

                                        = 0.387 J/g°C

Therefore, the specific heat capacity of copper is 0.387 J/g°C

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

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Explanation:

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R_{out} = \frac{A}{6000}*2000

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We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

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A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

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e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

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Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

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