a 191 g piece of copper is heated to 100 C in a boiling water bath and then dropped into a beaker containing 755 g of water (density =1.00 g/cm3) at 4.0 C. What is the final temperature of the copper and water after thermal equilibrium is reached? (Scu=0.385 J/g•K)
heat lost by Cu + heat gained by water = 0
(mass Cu x specific heat Cu x (Tfinal-Tinitial) + (mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
Substitute the numbers and solve for Tf.
To calculate the final temperature of the copper and water after thermal equilibrium is reached, we can use the principle of conservation of energy. The heat lost by the copper will be equal to the heat gained by the water.
First, let's calculate the heat lost by the copper. We can use the formula:
Q = m * c * ΔT
Where:
Q is the heat lost/gained (in Joules)
m is the mass (in grams)
c is the specific heat capacity (in J/g•K)
ΔT is the change in temperature (in °C)
The mass of the copper is given as 191 g, and the change in temperature is the difference between the initial temperature (100 °C) and the final temperature (let's call it Tf). So, the heat lost by the copper can be calculated as:
Qcopper = 191 g * 0.385 J/g•K * (100 °C - Tf)
Next, let's calculate the heat gained by the water. We can use the same formula:
Q = m * c * ΔT
The mass of the water is given as 755 g, and the change in temperature is the difference between Tf and the initial temperature of the water (4.0 °C). So, the heat gained by the water can be calculated as:
Qwater = 755 g * 1.00 J/g•K * (Tf - 4.0 °C)
Since heat lost by the copper is equal to the heat gained by the water, we can equate these two expressions:
191 g * 0.385 J/g•K * (100 °C - Tf) = 755 g * 1.00 J/g•K * (Tf - 4.0 °C)
Now we can solve this equation to find the value of Tf, which represents the final temperature of the copper and water at thermal equilibrium.