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1.3 kg of gold at 300 K comes in thermal contact with 2.4 kg of copper at 400 K. The specific heats of Au and Cu are 126 J/kg-K and 386 J/kg-K, respectively. What equilibrium temperature do they reach?

Answer :

Answer:

The final temperature of the metals will be 384.97 K

Explanation:

For the gold;

mass = 1.3 kg

temperature = 300 K

specific heat = 126 J/kg-K

For the copper;

mass = 2.4 kg

temperature = 400 K

specific heat = 386 J/kg-K

Firstly, we will have to calculate for the thermal energy possessed by each of the metal.

The heat possessed by a body = mcT

Where,

m is the mass of the body

c is the specific heat of the body, and

T is the temperature of the body at that instance

so we calculate for the thermal energy of the gold and the copper below

For gold;

heat energy = mcT = 1.3 x 126 x 300 = 49140 J

For copper;

heat energy = mcT = 2.4 x 386 x 400 = 370560 J

When the two metal come in thermal contact, this heat is evenly distributed between them.

The total heat energy = 49140 J + 370560 J = 419700 J

At thermal equilibrium, the two metals will be at the same temperature, to get this temperature, we equate the total thermal energy to the heat energy that will be possessed by the metals at equilibrium.

419700 = (1.3 x 126 x T) + (2.4 x 386 x T) = 163.8T + 926.4T

419700 = 1090.2T

T = 419700/1090.2 = 384.97 K

The final temperature of the metals will be 384.97 K

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Rewritten by : Barada

"The correct equilibrium temperature is [tex]$\boxed{T_{eq} = \frac{m_{Au}c_{Au}T_{Au,i} + m_{Cu}c_{Cu}T_{Cu,i}}{m_{Au}c_{Au} + m_{Cu}c_{Cu}}}$[/tex].

To find the equilibrium temperature [tex]$T_{eq}$[/tex] when the gold and copper come into thermal contact, we use the principle of conservation of energy. The heat lost by the hotter object (copper in this case) will be equal to the heat gained by the cooler object (gold).

The heat[tex]$Q$[/tex] absorbed or released by a substance can be calculated using the formula:

[tex]\[ Q = mc\Delta T \][/tex]

where[tex]$m$[/tex] is the mass of the substance, [tex]$c$[/tex] is the specific heat capacity of the substance, and [tex]$\Delta T$[/tex] is the change in temperature.

For gold (Au), we have:

- Mass,[tex]$m_{Au} = 1.3 \text{ kg}$[/tex]

- Specific heat capacity, [tex]$c_{Au} = 126 \text{ J/kg-K}$[/tex]

- Initial temperature, [tex]$T_{Au,i} = 300 \text{ K}$[/tex]

For copper (Cu), we have:

- Mass, [tex]$m_{Cu} = 2.4 \text{ kg}$[/tex]

- Specific heat capacity, [tex]$c_{Cu} = 386 \text{ J/kg-K}$[/tex]

- Initial temperature, [tex]$T_{Cu,i} = 400 \text{ K}$[/tex]

Since the heat lost by copper will be equal to the heat gained by gold, we can set up the equation:

[tex]\[ m_{Au}c_{Au}(T_{eq} - T_{Au,i}) = m_{Cu}c_{Cu}(T_{Cu,i} - T_{eq}) \] Now, we solve for $T_{eq}$[/tex]:

[tex]\[ m_{Au}c_{Au}T_{eq} - m_{Au}c_{Au}T_{Au,i} = m_{Cu}c_{Cu}T_{Cu,i} - m_{Cu}c_{Cu}T_{eq} \][/tex]

[tex]\[ m_{Au}c_{Au}T_{eq} + m_{Cu}c_{Cu}T_{eq} = m_{Au}c_{Au}T_{Au,i} + m_{Cu}c_{Cu}T_{Cu,i} \][/tex]

[tex]\[ T_{eq}(m_{Au}c_{Au} + m_{Cu}c_{Cu}) = m_{Au}c_{Au}T_{Au,i} + m_{Cu}c_{Cu}T_{Cu,i} \][/tex]

[tex]\[ T_{eq} = \frac{m_{Au}c_{Au}T_{Au,i} + m_{Cu}c_{Cu}T_{Cu,i}}{m_{Au}c_{Au} + m_{Cu}c_{Cu}} \][/tex]

Plugging in the values:

[tex]\[ T_{eq} = \frac{(1.3 \text{ kg})(126 \text{ J/kg-K})(300 \text{ K}) + (2.4 \text{ kg})(386 \text{ J/kg-K})(400 \text{ K})}{(1.3 \text{ kg})(126 \text{ J/kg-K}) + (2.4 \text{ kg})(386 \text{ J/kg-K})} \][/tex]