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How many joules are needed to warm 675 grams of water (specific heat= 4.186 J/g degrees Celsius) from 12 degrees Celsius to 85 degrees Celsius?

Sagot :

85-12 = 73 degrees needed
4.186 J/degree Celsius, so 
73 degrees * 4.186 J/degree = 305.578 J to raise 1 gram 73 degrees
there are 675 grams, so 305.578 * 675 = 206265.15 J

2.06 x 10^5 J are needed

Answer:

206,265.15 Joules are needed to warm 675 grams of water from 12 degrees Celsius to 85 degrees Celsius.

Explanation:

Between heat and temperature there is a direct proportionality relationship, where the constant of proportionality depends on the substance that constitutes the body. So, the equation that allows you to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged for a body of mass m, constituted by a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • Q=?
  • c=[tex]4.186 \frac{J}{g*degrees Celsius}[/tex]
  • m=675 g
  • ΔT=Tfinal-Tinitial=85 degrees Celsius - 12 degrees Celsius= 73 degrees Celsius.

Replacing:

[tex]Q=4.186 \frac{J}{g*degrees Celsius}*675 g*73 degrees Celsius[/tex]

Resolving you get:

Q=206,265.15 J

206,265.15 Joules are needed to warm 675 grams of water from 12 degrees Celsius to 85 degrees Celsius.