1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the
final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.

Question

21-07-2022
Physics

contestada

1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the
final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.

Respuesta :

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CintiaSalazar CintiaSalazar · Answer 1 · 2022-07-28T17:34:05+02:00

Taking into account the definition of calorimetry, the final temperature of the silver and water is 11.84 °C.

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.
ΔT is the temperature variation.

Final temperature of the silver and water

In this case, you know:

For silver:

Mass of silver= 500 g= 0.5 kg (being 1000 g= 1 kg)
Initial temperature of silver= 250 °C
Final temperature of silver= ?
Specific heat of silver= 240 [tex]\frac{J}{kgC}[/tex]

For water:

Mass of water = 1000 g= 1 kg
Initial temperature of water= 5 ºC
Final temperature of water= ?
Specific heat of water = 4180 [tex]\frac{J}{kgC}[/tex]

Replacing in the expression to calculate heat exchanges:

For silver: Qsilver= 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature of silver - 250 C)

For water: Qwater= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature of water - 5 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 silver gives up will be equal to the heat that the water receives. Therefore:

- Qsilver = + Qwater

And the final temperature of the silver is equal to the temperature of the water (Final temperature of silver= Final temperature of water= Final temperature). Then:

- 240 [tex]\frac{J}{kgC}[/tex]× 0.5 kg× (Final temperature - 250 C)= 4180 [tex]\frac{J}{kgC}[/tex] × 1 kg× (Final temperature - 5 C)

Solving:

- 120 [tex]\frac{J}{C}[/tex]× (Final temperature - 250 C)= 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)

120 [tex]\frac{J}{C}[/tex]× (250 C - Final temperature) = 4180 [tex]\frac{J}{C}[/tex]× (Final temperature - 5 C)

120 [tex]\frac{J}{C}[/tex]×250 C - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 4180 [tex]\frac{J}{C}[/tex]×5 C

30,000 J - 120 [tex]\frac{J}{C}[/tex]×Final temperature = 4180 [tex]\frac{J}{C}[/tex]×Final temperature - 20,900 J

50,900 J= 4300 [tex]\frac{J}{C}[/tex]×Final temperature

50,900 J÷ 4300 [tex]\frac{J}{C}[/tex]= Final temperature

11.84 °C= Final temperature

Finally, the final temperature of the silver and water is 11.84 °C.

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1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.

Respuesta :

Calorimetry

Final temperature of the silver and water

Otras preguntas

1. A 500 g piece of silver at 250°C is submerged in 1000 g of water at 5°C to be cooled. Determine the
final temperature of the silver and water. Given Cwater = 4180 J/kg°C and Csilver = 240 J/kg°C.