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  

c (diamond),

sio2 (quartz)

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.