Water (2450 g ) is heated until it just begins to boil. If the water absorbs 5.35×105 J of heat in the process, what was the initial temperature of the water?
Express your answer with the appropriate units.

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bharathparasad577 bharathparasad577 · Answer 1 · 2022-09-13T11:24:57+02:00

The initial temperature of water is 47.8⁰C

In this case, the challenge is to determine how much heat is required to bring a particular sample of water from its original temperature to its boiling point.

Given,

mass of water is 2450g

absorbed heat is 5.35×105 J

we have to find out initial temperature

First Calculate the change in temperature when 5.35×10^5 J of heat is provided to 2450g of water.

q = m . c . ∆T

q is amount of heat gained

m is the mass of water

c is the specific heat of the substance

∆T is change in temperature

Specific heat of water is 4.18 J/g⁰C

∆T = q / m . c

= 5.35×10^5 / 2450 × 4.18

= 52.2⁰C

Now we know that the final temperature of the water is

100⁰C and that this final temperature is 52.2⁰C

is higher than the initial temperature of the water.

So by, ∆T = T(f) - T(i)

T(i) = T(f) - ∆T

T(i) = 100⁰C - 52.2⁰C

= 47.8⁰C

Therefore, initial temperature of water is 47.8⁰C

Learn more about specific heat here:

https://brainly.com/question/27991746

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