Water (2350 g ) is heated until it just begins to boil. if the water absorbs 5.83×105 j of heat in the process, what was the initial temperature of the water? express your answer with the appropriate units.

40.7062 °C  

Explanation:

Let the initial temperature = x °C

Boiling temperature of water = 100 °C

Using,

Q = m C ×ΔT

Where,  

Q is the heat absorbed in the temperature change from x °C to 100 °C.

C gas is the specific heat of the water = 4.184 J/g  °C

m is the mass of water

ΔT = (100 - x) °C  

Given,

Mass = 2350 g

Q = 5.83 × 10⁵ J

Applying the values as:

Q = m C ×ΔT

5.83 × 10⁵ = 2350 × 4.184 × (100 - x)

x, Initial temperature = 40.7062 °C  

2b2t

explanation:

2b2t

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