A31.15−g stainless steel ball bearing at 103.27°c is placed in a constant-pressure calorimeter containing 110.3 g of water at 22.59°c. if the specific heat of the ball bearing is 0.474 j / (g · °c), calculate the final temperature of both the water and steel when they equilibrate. assume the calorimeter to have negligible heat capacity.

final temperature is  t = 19.92 °C

Explanation:

from the conservation of energy principle

Heat lost by  the steel ball is equal to the heat gained by water  and it is given as

Where

m = mass of steel ball = 31.15 g

c = specific heat of steel = 0.474 J / g -°C

\delta t = change in temperature of steel ball = 103.27°C  - t

where ,  t = final temperature of the system

m' = mass of water = 110.30 g

    c ' = specific heat of water = 4.186 J/g-°C  

= change in temperature of water = t-22.59

after putting all values we get

31.15*0.474*(103.27°C  - t) = 110.30*4.186*(t-22.59)

final temperature is  t = 19.92 °C

f=ma newton's second law 7x8=56n

overcharge is the right answer system