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A 10.0 g gold ring with a specific heat 0.129 at 24.00°C is placed in a calorimeter with 118 g of water at 1.00°C.
What will be the final temperature of the system?


Sagot :

Answer:

1.06 °C

Explanation:

From the question given above, the following data were obtained:

Mass of gold (M₉) = 10 g

Specific heat capacity of gold (C₉) = 0.129 J/gºC

Initial temperature of gold (T₉) = 24 °C

Mass of water (Mᵥᵥ) = 118 g

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Initial temperature of water (Tᵥᵥ) = 1 °C

Equilibrium temperature (Tₑ) =?

The equilibrium temperature of the system can be obtained as follow:

Heat loss by the gold = heat gained by the water

M₉C₉(T₉ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

10 × 0.129 (24 – Tₑ) = 118 × 4.184 (Tₑ – 1)

1.29(24 – Tₑ) = 493.712 (Tₑ – 1)

Clear bracket

30.96 – 1.29Tₑ = 493.712Tₑ – 493.712

Collect like terms

30.96 + 493.712 = 493.712Tₑ + 1.29Tₑ

524.672 = 495.002Tₑ

Divide both side by 495.002

Tₑ = 524.672 / 495.002

Tₑ = 1.06 °C

Therefore, the temperature of the system is 1.06 °C

The amount of heat of the system is measured by a device called a calorimeter. The final temperature of the system will be 1.06 degrees celsius.

What is equilibrium temperature?

The equilibrium temperature is the temperature that follows the law of thermodynamics and is said to be the system that has alike temperatures.  

Given,

Mass of Ag [tex]\rm (M_{g})[/tex] = 10g

Specific heat capacity of Ag [tex](\rm C_{g})[/tex] = [tex]\rm 0.129 J/g^{\circ}C[/tex]

The initial temperature of Ag [tex](\rm T_{g})[/tex] = [tex]24 ^{\circ}\;\rm C[/tex]

Mass of water [tex](\rm M_{w})[/tex] = 118 g

Specific heat capacity of water [tex](\rm C_{w})[/tex] = [tex]4.184 \rm \;J/g^{\circ}\;\rm C[/tex]

The initial temperature of water [tex](\rm T_{w})[/tex] = [tex]1 ^{\circ}\;\rm C[/tex]

Equilibrium temperature = [tex](\rm T_{e})[/tex]

The equilibrium temperature can be shown as, heat loss by the gold = heat gained by the water:

[tex]\rm \rm M_{g}C_{g}(T_{g} - T_{e}) = M_{w}C_{w}(T_{e}-C_{w})[/tex]

Substituting values in the equation:

[tex]\begin{aligned} 10 \times 0.129 (24 - \rm T_{e}) &= 118 \times 4.184 (\rm T_{e} - 1)\\\\\rm 1.29(24 - T_{e}) &= 493.712 (\rm T_{e} - 1)\\\\524.672 &= 495.002 \;\rm T_{e}\end{aligned}[/tex]

Now divide both the sides by 495.002:

[tex]\begin{aligned} \rm T_{e} &= \dfrac{524.672 }{495.002}\\\\\rm T_{e} &= 1.06 \;^{\circ}\rm C\end{aligned}[/tex]

Therefore, the final temperature of the system is 1.06 degrees celsius.

Learn more about equilibrium temperature here:

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