Question

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)
A. `80.7^(@)` C
B. `82.5^(@)` C
C. `83.2^(@)`C
D. `95.2^(@)`C

Answer 1

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