To warm 2.0 L of tea (d = 1.01 g/mL; specific heat = 4.184 J/g.°C) a cook places a 500. g block of stone (specific heat = 2.449J/g°C) at a temperature of 200 °F into the teapot. Assuming that the tea was initially at 72°F, what is the final temperature of the tea in °F?
Question
To warm 2.0 L of tea (d = 1.01 g/mL; specific heat = 4.184 J/g.°C) a cook places a 500. g block of stone (specific heat = 2.449J/g°C) at a temperature of 200 °F into the teapot. Assuming that the tea was initially at 72°F, what is the final temperature of the tea in °F?
Solution 1
To solve this problem, we need to use the concept of heat transfer. The heat gained by the tea is equal to the heat lost by the stone.
First, we need to convert the temperatures from Fahrenheit to Celsius because the specific heat is given in Celsius.
The formula to convert Fahrenheit to Celsius is: C = (F - 32) * 5/9
So, the initial temperature of the stone is (200-32)*5/9 = 93.33°C And the initial temperature of the tea is (72-32)*5/9 = 22.22°C
Next, we calculate the mass of the tea using the given volume and density. Mass = Volume * Density = 2.0L * 1.01g/mL = 2020g
Now, we can set up the equation for heat transfer.
The heat gained by the tea is equal to the heat lost by the stone.
m_tea * c_tea * (T_final - T_initial,tea) = m_stone * c_stone * (T_initial,stone - T_final)
Where: m_tea = mass of the tea = 2020g c_tea = specific heat of the tea = 4.184 J/g°C T_initial,tea = initial temperature of the tea = 22.22°C m_stone = mass of the stone = 500g c_stone = specific heat of the stone = 2.449 J/g°C T_initial,stone = initial temperature of the stone = 93.33°C T_final = final temperature of the tea and stone
Solving for T_final, we get:
T_final = (m_tea * c_tea * T_initial,tea + m_stone * c_stone * T_initial,stone) / (m_tea * c_tea + m_stone * c_stone)
Substituting the given values, we get:
T_final = (2020g * 4.184 J/g°C * 22.22°C + 500g * 2.449 J/g°C * 93.33°C) / (2020g * 4.184 J/g°C + 500g * 2.449 J/g°C)
Solving this, we get T_final = 26.67°C
Finally, we convert this back to Fahrenheit using the formula F = C * 9/5 + 32
So, the final temperature of the tea is 26.67°C * 9/5 + 32 = 80°F.
Solution 2
To solve this problem, we need to use the concept of heat transfer. The heat lost by the stone will be equal to the heat gained by the tea.
First, we need to convert the temperatures from Fahrenheit to Celsius because the specific heat is given in Celsius.
The formula to convert Fahrenheit to Celsius is: C = (F - 32) * 5/9
So, the initial temperature of the stone is (200-32)*5/9 = 93.33°C And the initial temperature of the tea is (72-32)*5/9 = 22.22°C
Next, we calculate the mass of the tea using the given volume and density. Mass = Volume * Density = 2.0L * 1.01g/mL = 2020g
Now, we can set up the equation for heat transfer.
The heat lost by the stone is equal to the heat gained by the tea.
m_stone * c_stone * (T_final - T_initial_stone) = m_tea * c_tea * (T_final - T_initial_tea)
Where: m_stone = mass of the stone = 500g c_stone = specific heat of the stone = 2.449J/g°C T_initial_stone = initial temperature of the stone = 93.33°C
m_tea = mass of the tea = 2020g c_tea = specific heat of the tea = 4.184J/g°C T_initial_tea = initial temperature of the tea = 22.22°C
We can solve this equation for T_final, the final temperature of the tea.
After solving the equation, we get the final temperature in Celsius. We then convert it back to Fahrenheit using the formula: F = C * 9/5 + 32
This will give us the final temperature of the tea in Fahrenheit.
Solution 3
To solve this problem, we need to use the concept of heat transfer. The heat gained by the tea is equal to the heat lost by the stone.
First, we need to convert the temperatures from Fahrenheit to Celsius because the specific heat is given in Celsius.
The formula to convert Fahrenheit to Celsius is: C = (F - 32) * 5/9
So, the initial temperature of the stone is (200-32)*5/9 = 93.33°C And the initial temperature of the tea is (72-32)*5/9 = 22.22°C
Next, we calculate the mass of the tea using the given volume and density. Mass = Volume * Density = 2.0L * 1.01g/mL = 2020g
Now, we can set up the equation for heat transfer.
The heat lost by the stone is equal to the heat gained by the tea.
m_stone * c_stone * (T_initial_stone - T_final) = m_tea * c_tea * (T_final - T_initial_tea)
Where: m_stone = mass of the stone = 500g c_stone = specific heat of the stone = 2.449J/g°C T_initial_stone = initial temperature of the stone = 93.33°C m_tea = mass of the tea = 2020g c_tea = specific heat of the tea = 4.184J/g°C T_initial_tea = initial temperature of the tea = 22.22°C T_final = final temperature of the tea and stone
Solving for T_final, we get:
T_final = (m_stone * c_stone * T_initial_stone + m_tea * c_tea * T_initial_tea) / (m_stone * c_stone + m_tea * c_tea)
Substituting the given values into the equation, we can calculate the final temperature in Celsius.
Finally, we convert the final temperature from Celsius back to Fahrenheit using the formula: F = C * 9/5 + 32
This will give us the final temperature of the tea in Fahrenheit.
Similar Questions
A 100 g glass container is at 10.0 °C. 200 g of water at 90.0 °C is added to the glass container. What is the final temperature of the water and the glass, in °C? (specific heat of water = 1.00 cal/g °C, specific heat of glass = 0.200 cal/g °C)
A 100 gram glass container contains 200 grams of water and 5 grams of ice all at 0 °C. A 200 gram piece of lead at 100 °C is added to the water and ice in the container. What is the final temperature of the system? (specific heat of ice = 2,000 J/kg °C , specific heat of water = 4,186 J/kg °C, heat of fusion of water = 333.7 kJ/kg, specific heat of glass = 837.2 J/kg °C, specific heat of lead = 127.7 J/kg °C)
0.595 g sample of steam at 107.4∘C is condensed into a container with 5.77g of water at 16.5∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J g⋅∘C, the specific heat of steam is 2.01 Jg⋅∘C, and Δ𝐻vap=40.7 kJ/mol.
A sample of 35.5 grams of liquid water is cooled from boiling point ( 100 °C). Approximately 3713 Joules of heat are released in this process. What is the final temperature of the liquid water? Liquid water has a specific heat of 4.184 J/g .
A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.67 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.