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Mrrafil [7]
3 years ago
15

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 d

egrees Celsius?
Chemistry
2 answers:
solong [7]3 years ago
5 0

Answer:

<u><em>206,265.15 Joules</em></u> 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=4.186 \frac{J}{g*degrees Celsius}
  • m=675 g
  • ΔT=Tfinal-Tinitial=85 degrees Celsius - 12 degrees Celsius= 73 degrees Celsius.

Replacing:

Q=4.186 \frac{J}{g*degrees Celsius}*675 g*73 degrees Celsius

Resolving you get:

Q=206,265.15 J

<u><em>206,265.15 Joules are needed to warm 675 grams of water from 12 degrees Celsius to 85 degrees Celsius.</em></u>

Naya [18.7K]3 years ago
3 0
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
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Answer:

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PV = nRT.................. Equation 1

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Substitute equation 2 into equation 1

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From the question, we were asked to find m.

There make m the subject of formula in equation 3

m = PVm'/RT.............. Equation 4

Given: P = 2.45 atm, V = 25.4 L, T = 482 K

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Bingel [31]

Here’s <em>one of many</em> possibilities.

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The reaction must supply 10.46 kJ.

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<em>n</em>Δ<em>H</em> = 10.46 kJ

<em>n</em> = 10.46 kJ/80 kJ = 0.131  mol

So, you need <em>0.131 mol A</em> and 0.131 mol B.

<em>Assume</em> you are using the <em>3 mol·L⁻¹ </em>solutions.

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You need 46.6 mL of 3 mol·L⁻¹ A + 46.6 mL of 3 mol·L⁻¹ B.

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