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FromTheMoon [43]
3 years ago
7

A student using this technique finds that when putting a hot piece of metal into a calorimeter which contains 35.070g of water,

the water temperature increases from 22.4 c to 24.7
c. how much heat (q) was absorbed by the water (s.h of water = 4.184 j/g k)
Chemistry
1 answer:
Aleks04 [339]3 years ago
5 0
Q=cmΔt,

c=<span>4.184 j/(g*⁰C)
m=35.070 g
</span>Δt=t(final)-t(initial) = 24.7-22.4=2.3⁰C
<span>
Q=4.184</span> j/(g*⁰C)* 35.070 g * 2.3⁰C ≈ 337.49 J ≈ 338 J
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a piece of food is burned in a calorimeter that contains 200.0g of water. If the temperature of the water rose from 65.0°C to 83
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Answer: 15062.4 Joules

Explanation:

The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of food = 200.0g

C = 4.184 j/g°C

Φ = (Final temperature - Initial temperature)

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Q = 200.0g x 4.184 j/g°C x 18°C

Q = 15062.4 J

Thus, 15062.4 joules of heat energy was contained in the food.

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3 years ago
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Answer:

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Answer:

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2 years ago
A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a
Andrew [12]

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

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You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

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For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

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That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

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Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

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Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

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Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

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