A 3.40 g piece of copper heated to 85ºC is immersed in 200 mL of water at 20ºC. What will be the final temperature of water after thermal equilibrium is reached? Ignore any heat losses. The specific heat capacity is 4.184 J/(g ºC) for water and 0.385 J/(g ºC) for copper.

To find the final temperature of the water when the copper is immersed in it, we can use the principle of conservation of energy. The heat gained by the water will be equal to the heat lost by the copper:

First, calculate the heat lost by copper using the formula:

[tex]\[Q_{\text{copper}} = m_{\text{copper}} \cdot c_{\text{copper}} \cdot \Delta T_{\text{copper}}\][/tex]

Where:

- [tex]\(m_{\text{copper}}\)[/tex] = Mass of copper = 3.40 g

-[tex]\(c_{\text{copper}}\)[/tex] = Specific heat capacity of copper = 0.385 J/(g°C)

- [tex]\(\Delta T_{\text{copper}}\)[/tex] = Change in temperature of copper = [tex]\(85°C - T_{\text{final}}\)[/tex]

Next, calculate the heat gained by the water using the formula:

[tex]\[Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\][/tex]

Where:

- [tex]\(m_{\text{water}}\)[/tex] = Mass of water = 200 g

- [tex]\(c_{\text{water}}\)[/tex] = Specific heat capacity of water = 4.184 J/(g°C)

- [tex]\(\Delta T_{\text{water}}\)[/tex] = Change in temperature of water =[tex]\(T_{\text{final}} - 20°C\)[/tex]

Since there is no heat loss, we can set [tex]\(Q_{\text{copper}}[/tex] = [tex]Q_{\text{water}}\)[/tex] and solve for [tex]\(T_{\text{final}}\)[/tex].

Now, we can calculate [tex]\(T_{\text{final}}\)[/tex] by rearranging the equations and solving for it. After the calculations, the final temperature of the water will be approximately 25.58°C.

Learn more about Thermal equilibrium

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