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KATRIN_1 [288]
3 years ago
9

The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary t

o increase the object's temperature from 6°C to 12°C? 4Q 2Q Q 3Q
Physics
2 answers:
Tresset [83]3 years ago
5 0

Answer:

The quantity of heat required to increase the temperature of the object from  6°C to 12°C is 3Q

Explanation:

Heat capacity is the quantity of heat required to increase the temperature of an object.

Q = mcΔθ

where;

Q is the quantity of heat

m is the mass of the object

c is specific heat capacity of the object

Δθ is change in temperature = T₂ - T₁

For the first sentence of this question;

Q = mc(6-4)

Q = mc(2)

Q = 2mc

For the second sentence of this question;

Let Q₂ be the quantity of heat required to increase the temperature of the object from  6°C to 12°C

Q₂ = mcΔθ

Q₂ = mc(12-6)

Q₂ = mc(6)

Q₂ = 6mc

Q₂ = 3(2mc)

Recall, Q = 2mc

Thus, Q₂ = 3Q

bixtya [17]3 years ago
4 0

Answer:

The answer is 3Q

Explanation:

The metal temperature increases in a linear way, we could get a difference between final and initial temperature

DT=FinalTemperature-InitialTemperature

We get a temperature difference of 2 degrees per each heat addition.  If we add the same heat 3 times more, it will increase to 12 degrees

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Calculate the equivalent resistance for both circuits. Series circuit: 2 Ω and 4 Ω Parallel circuit: 2 Ω and 4 Ω Which circuit h
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Equivalent resistance is also known as the overall resistance. 

For resistors in a series circuit, the total resistance is computed using the formula:

R_{T} = R_{1}+ R_{2}+ R_{3}... R_{n}

In other words, you just add up the resistance of each resistor in the series circuit. In your case you only have two resistors. You have 2Ω and 4Ω. So all you need to do is add that up. 

R_{T} = R_{1}+ R_{2}
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The total resistance of the series circuit is 6Ω

In a parallel circuit you get the total resistance using the formula:
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First you get the sum of all fractions and at the end take the reciprocal of the resulting fraction and divide. So let us take your problem into consideration where you have two resistors that have a resistance of 2Ω and 4Ω.

\frac{1}{R_{T}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}
\frac{1}{R_{T}} = \frac{1}{2}+\frac{1}{4}
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Get the reciprocal of the resulting fraction 3/4 and then divide. The reciprocal of 3/4 is 4/3.

4/3 = 1. 33Ω

So if you compare the equivalent resistance of the two circuits, the series circuit has a higher equivalent resistance. 
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