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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//g` K and that of water is `4.18 J//g ` K. 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. `9.5 J//K`B. `14 J//k`C. `95 J//K`D. `878 J//K` |
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Answer» Correct Answer - C (a) `250gmxx4.18xx(80-75)=(ms)(75-20)` `:.(ms)=95 J//k` (b)Granite (maximum specific heat) (c) as Rate `=500cal//min.:.5min=2500`cal. `2500xx4.18=(750xx4.18+500xx0.17)Deltatheta` Rightarrow `Deltatheta=3.24` final temperature`=83.24^(@)`C |
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