Question

The temperature inside a refrigerator is t2 °C and the room temperature is t, °C.
The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be ​

Answer 1

SOLUTION

GIVEN

The temperature inside a refrigerator is $\sf{{t_2 }^{ \circ} C}$ and the room temperature is $\sf{{t_1}^{ \circ} C}$

EVALUATION

The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

EVALUATION

The temperature inside a refrigerator is $\sf{{t_2 }^{ \circ} C}$ and the room temperature is $\sf{{t_1}^{ \circ} C}$

Now the temperature inside a refrigerator

$\sf{ = {t_2 }^{ \circ} C}$

$\sf{ = ({t_2 } + 273) \: K}$

$\sf{ =T_1}$

Again the room temperature

$\sf{ = {t_2 }^{ \circ} C}$

$\sf{ = ({t_2 } + 273) \: K}$

$\sf{ =T_2}$

The amount of heat given $\sf{ = Q_1}$

Let the amount of heat taken $= \sf{Q_2}$

Now it is given that the electrical energy consumed

= W = 1 joule

$\sf{ \therefore \: \: Q_2 = Q_1 - 1}$

Thus we have

$\displaystyle \sf{ \frac{Q_1}{Q_2} = \frac{T_1}{T_2} }$

$\displaystyle \sf{ \implies \frac{Q_1}{Q_1 - 1} = \frac{t_1 + 273}{t_2 + 273} }$

$\displaystyle \sf{ \implies \frac{Q_1 - 1}{Q_1 } = \frac{t_2 + 273}{t_1 + 273} }$

$\displaystyle \sf{ \implies 1 - \frac{ 1}{Q_1 } = \frac{t_2 + 273}{t_1 + 273} }$

$\displaystyle \sf{ \implies \frac{ 1}{Q_1 } =1 - \frac{t_2 + 273}{t_1 + 273} }$

$\displaystyle \sf{ \implies \frac{ 1}{Q_1 } = \frac{(t_1 + 273) - (t_2 + 273)}{t_1 + 273} }$

$\displaystyle \sf{ \implies \frac{ 1}{Q_1 } = \frac{(t_1 + 273 - t_2 - 273)}{t_1 + 273} }$

$\displaystyle \sf{ \implies \frac{ 1}{Q_1 } = \frac{(t_1 - t_2 )}{t_1 + 273} }$

$\displaystyle \sf{ \implies {Q_1 } = \frac{(t_1 + 273 )}{t_1 - t_2 } }$

FINAL ANSWER

The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

$\displaystyle \sf{ = \frac{(t_1 + 273 )}{t_1 - t_2 } }$

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