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. If, after some of the tea has been transferred to the thermos (as described in the passage), the teapot with its contents (at a temperature of 80^(@) C)was placed on the insulated warmer, what would be the temperature at the end of this 5 minute period?(Assume that no significant heat transfer occurs with the surroundings)

Answer»

`80.7^(@)` C
`82.5^(@)` C
`83.2^(@)`C
`95.2^(@)`C

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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