A42 kg sample of water absorbs 347 kj of heat. if the water was initially at 23.2 ∘c, what is its final temperature?

The  final  temperature  if  the  water  was  initially  at  23.3  degrees  is calculated  as  using  Q =Mc  delta  T
Q(heat)=   347 Kj
M=mass(42 Kg)
C=specific  heat  capacity  of  water(  4.187 Kj/Kg/K)
delta  T  is  the  change  in  temperature

347  Kj =  42  Kg x  4.187  Kj/ Kg/K  x  delta  T
347Kj=  175.854  Kj/K  x  delta  T
divide both  side  by  175.854 Kj k
347kj/175.854 kjk=175.854/175.854  x  delta  T
1.973  = delta T
if  the  initial  temperature  is  23.2  and  delta  T  is  1.973  therefore  final  temperature  =  1.973  +23.2=25.173 degrees 

Whenever we deal with the problems for checking the amount of heat lost or gained(absorbed) by the fluid, we use the following equation:

--- (A)

Where,
Q = The amount of heat lost or gained(absorbed) by the fluid
m = Mass of the fluid
= Heat Capacity of the fluid
= Change in Temperature = Final Temperature() - Initial Temperature()

Data given:
Initial Temperature = = 23.2°C
The amount of heat absorbed by the water = 347 kJ
Mass of the water = m = 42 kg
Heat Capacity of water = = 4.184 kJ/kg°C
Final Temperature =  = ?

Plug-in the values in equation (A)

(A) => 347 = 42 * 4.184 * ( - 23.2)

=>  ≈ 25.175°C

Ans: The final temperature of water ≈ 25.175°C

-i

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.