Oh no! Billy forgot that his calorimeter should have 2 coffee cups to contain all the heat. If Billy dropped a 25.0 g piece of solid iron (Cs= 0.449 J/g°C) at 398 K in a coffee cup containing 25.0 mL of water at 298K and because of his mistake, 21% of the heat that should be transferredfrom the iron to the wateris lost to the atmosphere, calculate the final temperature of the iron water mixture?

Explanation:

Let the final temperature be T .

heat lost by iron = mass x specific heat x decrease in temperature

= 25 x .449 x ( 398 - T  )

Heat applied to increase the temperature of water

= 25 x .449 x ( 398- T  )  x .79  [ 21 % is lost ]

= 8.86775 ( 398 - T  )

heat gained by water = 25 x 4.2 x ( T - 298 )

Both of them are equal

8.86775 ( 398 - T  )  = 25 x 4.2 x ( T - 298 )

8.86775 ( 398 - T  )  = 105 x ( T - 298 )

105 T + 8.86775 T = 105 x 298 + 398 x 8.86775

113.86775 T = 31290 + 3529.36

113.86775 T = 34819.36

T = 305.78 K .

ch3 [ch2]14cooh

explanation:

answer; /// i do believe that the correct answer is (c)nuclear bond