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Water (2450 g ) is heated until it just begins to boil. If the water absorbs 5.61×105 J of heat in the process, what was the initial temperature of the water?

Express your answer with the appropriate units.

Respuesta :

Answer:

45.3 °C was the initial temperature

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 = 5.61 x 10^5 J

We will use the formule : Q = mcΔT

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with ΔT = change in temperature ( in °C or K)

The  specific heat of water is 4.186 J/g °C

Step 2 : Calculate the initial temperature

To find the initial temperature, we have to rearrange the formule:

ΔT = Q / mc

In this case we have :

ΔT = 5.61 * 10^5 J / 2450 g * 4.186 J/g °C = 54.70

⇒ The final temperature of the water is the boiling point (100 °C). The change of temperature is 54.70  This means that the boiling point is 54.70°C higher than the initial temperature.

This means : ΔT = Tboiling point - Tinitial

ΔT = 54.70 °C = 100 °C - Tinitial

Tinitial = 100 °C - 54.70 °C = 45.3 °C

The initial temperature of the water is 45.3 °C

BABYSUNFLOWER8
42.9 C calculate the change in temperature that occurs when you provide
5.47
⋅
10
5
J
of heat to
2290 g
of water.

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