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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