A 500 g teapot and an insulated thermos are in a 20^(@)C room. The teapot is filled with 1000 g of the boiling water. 12 tea bags are then placed into the teapot. The brewed tea is allowed to cool to 80^(@)C, then 250 g of the tea is poured from the teapot into the thermos. The teapot is then kept on an insulated warmer that transfers 500 cal//min to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature. The specific heat of teapot is 0.17 J//gK and that of water is 4.18 J//gK. The entire procedure is done under atmospheric pressure. There are 4.18 J in one calorie. After the tea is added to the thermos, the temperature of the liquid quickly falls from 80^(@)C to 75^(@)C as it reaches thermal equilibrium with the thermos flask. What is the heat capacity of the thermos?

A 500 g teapot and an insulated thermos are in a 20^(@)C room. The tea

1.	A 500 g teapot and an insulated thermos are in a 20^(@)C room. The teapot is filled with 1000 g of the boiling water. 12 tea bags are then placed into the teapot. The brewed tea is allowed to cool to 80^(@)C, then 250 g of the tea is poured from the teapot into the thermos. The teapot is then kept on an insulated warmer that transfers 500 cal//min to the tea. Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature. The specific heat of teapot is 0.17 J//gK and that of water is 4.18 J//gK. The entire procedure is done under atmospheric pressure. There are 4.18 J in one calorie. After the tea is added to the thermos, the temperature of the liquid quickly falls from 80^(@)C to 75^(@)C as it reaches thermal equilibrium with the thermos flask. What is the heat capacity of the thermos?
Answer» <html><body><p>`9.5 J//K`<br/>`14 J//k`<br/>`95 J//K`<br/>`878 J//K`</p>Solution :(a) `250gmxx4.18xx(80-75)=(<a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>)(75-20)` <br/> `:.(ms)=95 J//k` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)Granite (maximum <a href="https://interviewquestions.tuteehub.com/tag/specific-1220917" style="font-weight:bold;" target="_blank" title="Click to know more about SPECIFIC">SPECIFIC</a> heat) <br/> (c) as Rate `=500cal//min.:.5min=2500`cal. <br/> `2500xx4.18=(750xx4.18+500xx0.17)Deltatheta` <br/> <a href="https://interviewquestions.tuteehub.com/tag/rightarrow-623803" style="font-weight:bold;" target="_blank" title="Click to know more about RIGHTARROW">RIGHTARROW</a> `Deltatheta=3.24` final temperature`=83.24^(@)`C</body></html>

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