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14.0 J of heat is added to 10.00 g of gold, which is initially at 25°C.

The heat capacity of gold is 25.41 J/(mol • °C).

What would be the final temperature of the gold?

Answer :

Answer:

36°C is your answer

Explanation:

use the molar mass to find moles:

(10.00g Au)(1 mole Au/197 grams)= 0.05076 moles of gold

dH= n C dT

14.0 J = (.05076 moles) (25.41 J/mol* degree C) (dT)

dT=10.85 celsius rise in temp

if initially at 25 degrees celsius, a 10.85 celsius rise in temp will bring it to 35.85.

so round that answer and you should get 36 degrees as your answer

(hope this helped :) )

CintiaSalazar

Considering the definition of calorimetry, the final temperature of the gold is 35.85 °C.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In other words, calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.

The amount of heat that a body receives or transmits is determined by the following expression:

Q = c×m×ΔT

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

In this case, you know:

  • Q= 14 J
  • c= 25.41 [tex]\frac{J}{molC}[/tex]
  • m= 10 g×[tex]\frac{1 mol}{197 grams}[/tex]= 0.05076 moles (where 197 [tex]\frac{grams}{mol}[/tex] is the molar mass of gold)
  • ΔT= Tfinal - Tinitial= Tfinal - 25 C

Then:

14 J= 25.41 [tex]\frac{J}{molC}[/tex]× 0.05076 moles× (Tfinal - 25 C)

Solving:

[tex]\frac{14 J}{ 25.41 \frac{J}{molC}x 0.05076 moles} =Tfinal - 25 C[/tex]

[tex]Tfinal=\frac{14 J}{ 25.41 \frac{J}{molC}x 0.05076 moles} + 25 C[/tex]

Tfinal= 35.85 °C

Finally, the final temperature of the gold is 35.85 °C.

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  • brainly.com/question/11586486?referrer=searchResults
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