Water 2270g is heated until it just begins to boil. if the water absorbs 5.51 ×10^5 j of heat in the process, what was the initial temperature of the water?

42.01 °C

Explanation:

Step 1: explain the problem

We have to find the initial temperature, when a certain amount of heat raises this sample of water to its boiling point ( 100 °C)  

⇒this amount of heat = 5.51 x 10Δ^5 J

We will use the formule : Q = mcΔ T

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with Δ T = temperature change ( in °C or K)

Water has a specific heat of 4.186 J/g °C

Step 2 : Calculate the initial temperature

We have to rearrange the formule first:

Δ T = Q / mc

In this case we have :

Δ T = 5.51 * 10^5 J / 2270 g * 4.186 J/g °C = 57.99

⇒ The final temperature of the water is  the boiling point (100 °C) and the change of temperature is 57.99. This means that the boiling point is 57.99 °C higher than the initial temperature.

This means : Δ T = Tboiling point - Tinitial

Δ T = 57.99 °C = 100 °C - Tinitial

Tinitial = 100 °C - 57.99 °C = 42.01 °C

The initial temperature of the water is 42.01 °C

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definition

                    entropy is a measure of the randomness or the extent of disorder of a chemical process.

                      enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

                        entropy is measured in joule per kelvin (jk⁻¹)   whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

                            entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

                              in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

sup bro, im dumb, but this is what i found xd

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