Question

Efficiency of carnot cycle changes from 1/6 to 1/3 when source temperature is raised by 100k calculate the temperature of sink

Answer 1

T2= y (in both cases)
therefore,
the resulting
x = -700
therefore, from eq.1
Efficiency = 1-T2/T1
1/6 = 1 - y/-700
therefore
y= -583.3

Answer 2

Answer:

Temperature of sink is 333.33 K.

Explanation:

• A cannot engine working between source temperature, T1 and sink temperature, T2.
• The efficiency of the cannot engine is given by

$η = 1- \frac{T2}{T1}$

Given that :

• Efficiency changes from 1/6 to 1/3 when source temperature is raised by 100K

To find :

• Temperature of sink

Solution :

• Let, the source temperature of carnot Engine be T1 and sink temperature be T2.
• Efficiency of engine, η1 is 1/6.

$η1= 1- \frac{T2}{T1}\\ \frac{T2}{T1}= 1-\frac{1}{6} \\ \frac{T2}{T1}= \frac{5}{6} \\ T2 = \frac{5}{6}T1 \\ T1=\frac{6}{5}T1 ---(1)$

• Now, temperature of source is increased by 100K and new efficiency be 1/3.

$η2= 1- \frac{T2}{T1+100}\\ \frac{T2}{T1+100}= 1-\frac{1}{3} \\ \frac{T2}{T1+100}= \frac{2}{3} \\ T2 = \frac{2}{3}(T1+100) \\ T2 =\frac{2}{3}( \frac{6}{5} T2 + 100) \\ T2 = \frac{4}{5} T2 + \frac{200}{3} \\ \frac{1}{5} T2 = \frac{200}{3} \\ T2 = \frac{1000}{3} = 333.33 \: k$

• Hence, temperature of sink is 333.33 K .