Question

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?

Answer 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