Water (2290 g) is heated until it just begins to boil. If the water absorbs 5.47×105J of heat in the process what was the initial temperature of the water?
Answers
Answered by
1
Answer:
50.952 °C or 323.952 K
Explanation:
Mass of water = 2290 g = 2.29 Kg
Heat absorbed (Q) = 5.47 x 10⁵ J = 547 KJ
Since water is boiling, final temperature of water = 100°C = 373 K
when water is getting warmed, the amount of heat absorbed to increase the temperature of water by 1°C is given by heat capacity.
Specific heat capacity of water = 4.87 KJ/Kg
Mathematically,
Q = mCΔT
where
Q = Heat absorbed
m= mass of the water
C = specific heat capacity of water
ΔT = Temperature change = Final temperature - initial temperature
Putting numerical values
547 = 2.29 x 4.87 x ΔT
⇒ΔT = 49.048
⇒Final temperature - Initial temperature = 49.048°C
⇒100°C - Initial temperature = 49.048°C
⇒ Initial temperature = 100 - 49.048 = 50.952°C
Similar questions