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4 years ago

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If you are given an unknown liquid that is 1.0 L and has the mass of 500 grams which of the substance would it be. Distilled Wat

er Density= 1.0g/cm^3, Propane density 0.494 g/cm^3, Salt Water density 1.025 g/cm^3 or Liquid Gold 17.31 g/cm^3?

1 answer:

kodGreya [7K]4 years ago

8 0

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Two aircraft rivets, one iron and the other copper, are placed in a calorimeter that has an initial temperature of 20.°C. The da

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The heat will flow from Cu to Fe because Cu has higher temperature and has lower peak value .

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two aircraft rivets , one iron and the other copper , are placed in a calorimeter .

the initial temp of the calorimeter is 20 degC .

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A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.

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Answer:

<u>Question 1: 0.2J/(gºC)</u>

<u>Question 2: 6,000J</u>

<u>Question 3: 300J</u>

<u>Question 4: 80g</u>

<u>Question 5: 74ºC</u>

<u>Question 6: 50g</u>

<u></u>

Explanation:

Question 1.<em> A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.</em>

<em />

The thermal energy equation is:

Q = m × C × ΔT

<em />

Substitute and solve for C:

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<em />

You must round to one significant figure because one factor has one significant figure).

<em />

<em />

Qustion 2.<em> 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)</em>

<em />

Use the thermal energy equation again:

Q = m × C × ΔT

<em />

Substitute and compute:

Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)

Q = 5,857.6J

Round to one significant figure: 6,000J

<em />

Question 3. <em>Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?</em>

Q = m × C × ΔT

Q = 50g × 0.420J/gºC × (35ºC - 50ºC)

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Round to one significant figure (because 50g has one significant figure)

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<em />

Question 4.<em> </em><em>Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?</em>

<em />

Q = m × C × ΔT

Solve for m:

m = Q / (C × ΔT)

Substitute and compute:

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<em />

Q = m × C × ΔT

<em />

Q is negative, since it is released.

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T = - 7500J / 400g × 4.184J/gºC) + 70ºC

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<em />

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

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m = 9,500J / (4.184J/gºC × 50ºC)

m = 45g

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m = 50g.

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