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A 23.9 g sample of iridium is heated to 89.7°C, and then dropped into 20.0 g of water in a foam-cup calorimeter. The temperature of the water went from 20.1°C to 22.6°C. Calculate the specific heat of iridium (specific heat of water = 4.18 J/g.°C)

Respuesta :

Answer: The specific heat of iridium is 0.130 J/g°C

Explanation:

When iridium is dipped in water, the amount of heat released by iridium will be equal to the amount of heat absorbed by water.

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]      ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of iridium = 23.9 g

[tex]m_2[/tex] = mass of water = 20.0 g

[tex]T_{final}[/tex] = final temperature = 22.6°C

[tex]T_1[/tex] = initial temperature of iridium = 89.7°C

[tex]T_2[/tex] = initial temperature of water = 20.1°C

[tex]c_1[/tex] = specific heat of iridium = ?

[tex]c_2[/tex] = specific heat of water = 4.18 J/g°C

Putting values in equation 1, we get:

[tex]23.9\times c_1\times (22.6-89.7)=-[20\times 4.18\times (22.6-20.1)][/tex]

[tex]c_1=0.130J/g^oC[/tex]

Hence, the specific heat of iridium is 0.130 J/g°C

CintiaSalazar

Taking into account the definition of calorimetry, the specific heat of iridium is 0.13 [tex]\frac{J}{gC}[/tex].

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  specific heat substance.
  • ΔT is the temperature variation.

Specific heat of iridium

In this case, you know:

  • For iridium:
  1. Mass of iridium = 13.5 g
  2. Initial temperature of gold= 89.3 °C
  3. Final temperature of gold= 22.6 ºC
  4. Specific heat of gold = Unknown
  • For water:
  1. Mass of water = 20 g
  2. Initial temperature of water= 20.1 ºC
  3. Final temperature of water= 22.6 ºC
  4. Specific heat of water = 4.18 [tex]\frac{J}{gC}[/tex]

Replacing in the expression to calculate heat exchanges:

For iridium: Qiridium= c × 23.9 g× (22.6 C - 89.7 C)

For water: Qwater= 4.186 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 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:

- Qiridium = + Qwater

- c × 23.9 g× (22.6 C - 89.7 C)= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

Solving:

c × 23.9 g× ( 89.7 C - 22.6 C)= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

c × 23.9 g× 67.1 C= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× 2.5 C

c × 1603.69 g×C= 209 J

[tex]c=\frac{209 J}{1603.69 gC}[/tex]

c= 0.13 [tex]\frac{J}{gC}[/tex]

Finally, the specific heat of iridium is 0.13 [tex]\frac{J}{gC}[/tex].

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