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Select the correct answer from each drop-down menu. a 126.1-gram block of granite at 92.6°c is dropped into a tub of water at 24.7°c in an isolated system. the final temperature of both the granite and the water is 51.9°c. the specific heat capacity of granite is 0.795 joules/gram degree celsius, and the specific heat capacity of water is 4.186 joules/gram degree celsius. the granite block transferred of energy, and the mass of the water is .

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CintiaSalazar

Taking into account the definition of calorimetry, the mass of water is 35.70 grams.

What is calorimetry

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

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • c is the specific heat substance c.
  • ΔT is the temperature variation.

Mass of water in this case

In this case, you know:

For granite:

  • Mass of granite= 126.1 g
  • Initial temperature of granite= 92.6 °C
  • Final temperature of granite= 51.9 ºC
  • Specific heat of granite = 0.795 [tex]\frac{J}{gC}[/tex]

For water:

  • Mass of water = ?
  • Initial temperature of water= 24.7 ºC
  • Final temperature of water= 51.9 ºC
  • Specific heat of water = 4.186 [tex]\frac{J}{gC}[/tex]

Replacing in the expression to calculate heat exchanges:

For gold: Qgranite= 0.795 [tex]\frac{J}{gC}[/tex]  × 126.1 g× (51.9 °C - 92.6 °C)

For water: Qwater= 4.186 [tex]\frac{J}{gC}[/tex]× mass of water× (51.9 °C - 24.7 °C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qgranite = + Qwater

- 0.795 [tex]\frac{J}{gC}[/tex]  × 126.1 g× (51.9 °C - 92.6 °C)= 4.186 [tex]\frac{J}{gC}[/tex]× mass of water× (51.9 °C - 24.7 °C)

Solving:

4,080.15465 J= 4.186 [tex]\frac{J}{gC}[/tex]× mass of water× 27.3 °C

4,080.15465 J= 114.2778 [tex]\frac{J}{g}[/tex]× mass of water

mass of water= (4,080.15465 J)÷ 114.2778 [tex]\frac{J}{g}[/tex]

mass of water= 35.70 grams

Finally, the mass of water is 35.70 grams.

Learn more about calorimetry:

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