Question

a carnot engine whose source at 1000k and takes 800 cal heat and rejects600calheat what is temp ofsink
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Answer 1

Temperature of the source = (10)³ k

Heat taken up from the source = 800 cal

Heat rejected to the sink = 600 cal

We know that ,

$\underline{ \large \mathtt{ \fbox{ \frac{ T_{2}}{T_{1}} = \frac{Q_{2}}{Q_{1}} }}}$

Where ,

T1 is the temperature of the source

T2 is the temperature of the sink

Q1 is the heat taken up from the source

Q2 is the heat rejected to the sink

Substitute the known values , we get

$\sf \hookrightarrow \frac{T_{2}}{1000} = \frac{800}{600} \\ \\ \sf \: \: By \: cross \: multiplication \: , \: we \: get \\ \\ \sf \hookrightarrow T_{2} = \frac{8 \times {(10)}^{3} }{6} \\ \\ \sf \hookrightarrow T_{2} = \frac{6 \times {(10)}^{3} }{8} \\ \\ \sf \hookrightarrow T_{2} =1.33 \times {(10)}^{3} \: \: k$

Hence , the temperature of the sink is 1.33 × (10)³ k

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Answer 2

The temperature of the sink is 750K.

Given:

A Carnot engine whose source is at 1000k takes 800 cal heat and rejects 600cal heat.

To Find:

The temperature of the sink.

Solution:

To find the temperature of the sink we will follow the following steps:

As we know,

Carnot engine is used to attain maximum efficiency by converting heat into work and operating between two temperatures.

The formula relating heat absorbed, rejected and the temperature of the source and sink is:

$\frac{t2}{t1} = \frac{q2}{q1}$

t2 is the temperature of the sink and t1 is the temperature of the source.

q1 and q2 are the heat absorbed and rejected respectively.

So,

According to the question:

q1 = 800 cal

q2 = 600 cal

t1 = 1000K

t2 =?

Now,

Putting values in the above formula we get,

$\frac{t2}{1000} = \frac{600}{800}$

$t2 = \frac{8}{6} \times 1000 = 750k$

Henceforth, The temperature of the sink is 750K.

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