Question

A 42 kg sample of water absorbs 347 kJ of heat. If the water was initially at 23.2 ∘C, what is its final temperature?

Answer 1

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


`Q = m * C_(p) * dT` --- (A)

Where,
Q = The amount of heat lost or gained(absorbed) by the fluid
m = Mass of the fluid

`C_(p)` = Heat Capacity of the fluid

`dT` = Change in Temperature = Final Temperature(
`T_(f)`) - Initial Temperature(
`T_(i)`)

Data given:
Initial Temperature =
`T_(i)` = 23.2°C
The amount of heat absorbed by the water = 347 kJ
Mass of the water = m = 42 kg
Heat Capacity of water =
`C_(p)` = 4.184 kJ/kg°C
Final Temperature =
`T_(f)` = ?

Plug-in the values in equation (A)

(A) => 347 = 42 * 4.184 * (
`T_(f)` - 23.2)

=>
`T_(f)` ≈ 25.175°C

Ans: The final temperature of water ≈ 25.175°C

-i

Answer 2

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