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sweet [91]
3 years ago
9

A copper block ( 90°C) is placed in contact with a lead block ( 20°C), which is already in contact with an iron block (70°C). Wh

ich of the following will happen?
A. Heat will flow from the copper to the lead and from the iron to the lead until the temperatures are equal .

B. Heat will flow from the iron to the lead to the copper until the temperatures are equal.

C. Heat will flow from the copper to the lead to the iron until the temperatures are equal.

D. Heat will flow from the lead to the copper and to the iron until the temperatures are equal.
Chemistry
2 answers:
Ostrovityanka [42]3 years ago
3 0
<span>B. Heat will flow from the iron to the lead to the copper until the temperatures are equal. because i don't really speak Italian </span><span>
</span>
Radda [10]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

According to zeroth law of thermodynamics, when two systems are in thermal equilibrium with the third system then all of them are in thermal equilibrium with each other.

Hence, when copper block is placed adjacent to lead block and lead block is placed adjacent to iron block then heat will flow from hotter block to the colder block.

Thus, we can conclude that heat will flow from the copper to the lead and from the iron to the lead until the temperatures are equal.

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Explanation:

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How much heat is required to change temperature of 10 g of water from 4 °C to 8 °C? (Water has a specific heat of 4.18 )?
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Explanation:

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Identify which subatomic particles match each of these descriptions. One of the descriptions describes two particles. Make sure
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3 years ago
the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
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<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

n = moles of acetone = 6.3 moles

\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

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4 0
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