Question

please answer this I will mark as brainliest it's urgent please ​

Answer 1

Answer:

Here is your answer I hope this is help full

Answer 2

Answer:

• Temperature (T) of the Source is 480 K.

Given:

CASE-1

1. Efficiency of Carnot Engine (η) = 40 %.
2. Temperature of Source (T₂) = 400 K.

CASE-2

1. Efficiency of Carnot Engine (η) = 50 %.

Explanation:

$\rule{300}{1.5}$

From The Carnot Engine's Formula we Know,

$\large\bigstar \: {\boxed{\tt \eta = 1 - \dfrac{T_2}{T_1}}}$

$\bold{Here}\begin{cases}\eta \: \text{Denotes Efficiency} \\ T_2 \: \text{Denotes Temperature of Exhaust} \\ T_1 \; \text{Denotes Temperature of Source}\end{cases}$

Now,

$\large{\boxed{\tt \eta = 1 - \dfrac{T_2}{T_1}}}$

Substituting the values,

$\large{\tt \longmapsto 40 \% = 1 - \dfrac{T_2}{400}}$

$\large{\tt \longmapsto \dfrac{40}{100} = 1 - \dfrac{T_2}{400}}$

$\large{\tt \longmapsto \cancel{\dfrac{40}{100}} = 1 - \dfrac{T_2}{400}}$

$\large{\tt \longmapsto \dfrac{4}{10} = 1 - \dfrac{T_2}{400}}$

$\large{\tt \longmapsto \dfrac{T_2}{400} = 1 - \dfrac{4}{10}}$

$\large{\tt \longmapsto \dfrac{T_2}{400} = \dfrac{10 - 4}{10}}$

$\large{\tt \longmapsto \dfrac{T_2}{400} = \dfrac{6}{10}}$

$\large{\tt \longmapsto T_2 = \dfrac{6}{10} \times 400}$

$\large{\tt \longmapsto T_2 = \dfrac{6}{\cancel{10}} \times \cancel{400}}$

$\large{\tt \longmapsto T_2 = 6 \times 40}$

$\large\longmapsto{\underline{\boxed{\tt T_2 = 240 \: K}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, this Exhaust Temperature is same For Second case.

From The Carnot Engine's Formula we Know,

$\large\bigstar \: {\boxed{\tt \eta = 1 - \dfrac{T_2}{T_1}}}$

$\bold{Here}\begin{cases}\eta \: \text{Denotes Efficiency} \\ T_2 \: \text{Denotes Temperature of Exhaust} \\ T_1 \; \text{Denotes Temperature of Source}\end{cases}$

Now,

$\large{\boxed{\tt \eta = 1 - \dfrac{T_2}{T_1}}}$

Substituting the values,

$\large{\tt \longmapsto 50 \% = 1 - \dfrac{240}{T_1}}$

$\large{\tt \longmapsto \dfrac{50}{100} = 1 - \dfrac{240}{T_1}}$

$\large{\tt \longmapsto \cancel{\dfrac{50}{100}} = 1 - \dfrac{240}{T_1}}$

$\large{\tt \longmapsto \dfrac{5}{10} = 1 - \dfrac{240}{T_1}}$

$\large{\tt \longmapsto \dfrac{240}{T_1} = 1 - \dfrac{5}{10}}$

$\large{\tt \longmapsto \dfrac{240}{T_1} = \dfrac{10 - 5}{10}}$

$\large{\tt \longmapsto \dfrac{240}{T_1} = \dfrac{5}{10}}$

$\large{\tt \longmapsto \dfrac{240}{T_1} = \cancel{\dfrac{5}{10}}}$

$\large{\tt \longmapsto \dfrac{240}{T_1} = \dfrac{1}{2}}$

$\large{\tt \longmapsto T_1 \times 1 = 240 \times 2}$

$\large{\tt \longmapsto T_1 = 240 \times 2}$

$\huge{\boxed{\boxed{\tt T_1 = 240 \: K}}}$

∴ Temperature (T) of the Source is 480 K (Option- 2).

$\rule{300}{1.5}$