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What is the total number of Joules lost when 10. grams of water at 80.°C is cooled to 60.°C?

Sagot :

Sarah06109

Answer:

[tex]\boxed {\boxed {\sf 836.8 \ Joules \ lost}}[/tex]

Explanation:

Since we are given the mass and change in temperature, we should use the following formula for heat.

[tex]q= mc \Delta T[/tex]

In this formula, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

We know the mass of the water is 10 grams. We will have to look up water's specific heat, because it is not given. It is 4.184 J/g °C.

Let's find the change in temperature. This is the difference of the final and initial temperature. The water started at 80 °C and cooled to 60°C.

  • ΔT= final temperature - initial temperature
  • ΔT= 60° C - 80°C
  • ΔT= -20 °C

Now we have values for each variable.

  • m= 10 g
  • c= 4.184 J/g°C
  • ΔT= -20°C

Substitute the values into the formula.

[tex]q= 10 \ g * 4.184 \ J/g \textdegree C * -20 \textdegree C[/tex]

Multiply the first 2 numbers. The units of grams will cancel each other out.

[tex]q= 41.84 \ J/ \textdegree C * -20 \textdegree C[/tex]

Multiply again. This time, the degrees Celsius cancel.

[tex]q= -836.8 \ J[/tex]

A total of 836.8 Joules are lost.

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