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. An alternative method for keeping the tea hot would be, to place the teapot on a 10 pound block that has been heated in an oven to 300^(@)C. A block of which of the following substances would best be able to keep the tea hot?

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//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. An alternative method for keeping the tea hot would be, to place the teapot on a 10 pound block that has been heated in an oven to 300^(@)C. A block of which of the following substances would best be able to keep the tea hot?
Answer» copper (specific heat `=0.39 J//gK)` GRANITE (specific heat`=0.79 J//gK)` iron (specific heat `=0.45 J//g K)` pewter (specific heat `=0.17 J//g K)` SOLUTION :(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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