Water (3090 g ) is heated until it just begins to boil. if the water absorbs 5.09×105 j of heat in the process, what was the initial temperature of the water?

Respuesta :

The initial temperature of the water was 60.64 degree Celsius.

What is heat energy?

Heat is the type of energy that travels from a high-temperature system to a low-temperature system when it is transmitted between things or systems of different temperatures. also known as thermal energy or heat energy. Usually, heat is expressed in Btu, calories, or joules.

Calculation for the initial temperature of the water;

The formula for release of heat energy is given by;

Q = mc∆t

Q = heat energy

m = mass

c = specific heat (of water)

∆t= change in temperature (= final - initial temp)

final temp of water = boiling point of water i.e. 100*C

'c' for water = 4.186 joule/gram °C

The mass 'm' = 3090 g.

The energy absorbed by water is 5.09×[tex]10^{5}[/tex] = 509000 J.

Let 'T' be the initial temperature.

Substitute all the value in formula to get the equation.

509000 =  3090 × 4.186 × (100 - T)

100 - T = 39.35

T = 100 - 39.35

T = 60.64

Therefore, the initial temperature of the water was 60.64 degree Celsius.

To know more about the thermal energy, here

https://brainly.com/question/19666326

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