A piece of tungsten of mass 50 g is placed over a flame for long time. The metal is then quickly transferred to a well-insulated aluminum calorimeter of mass 120 g containing 300 g of water at 22 °C. After some time the temperature of the water reaches a maximum value of 31 °C. You may use these values of specific heat capacity : water 4.2 × 103 J kg−1 K−1; tungsten 1.05 × 102 J kg−1 K−1; aluminium 9.45 × 102 J kg−1 K−1. What is the temperature of the flame?
Question
A piece of tungsten of mass 50 g is placed over a flame for long time. The metal is then quickly transferred to a well-insulated aluminum calorimeter of mass 120 g containing 300 g of water at 22 °C. After some time the temperature of the water reaches a maximum value of 31 °C. You may use these values of specific heat capacity : water 4.2 × 103 J kg−1 K−1; tungsten 1.05 × 102 J kg−1 K−1; aluminium 9.45 × 102 J kg−1 K−1. What is the temperature of the flame?
Solution
To find the temperature of the flame, we can use the principle of conservation of energy.
First, let's calculate the heat gained by the water in the calorimeter. We can use the formula:
Q = mcΔT
Where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given: m_water = 300 g = 0.3 kg c_water = 4.2 × 10^3 J kg^−1 K^−1 ΔT_water = 31 °C - 22 °C = 9 °C
Using the formula, we have: Q_water = (0.3 kg)(4.2 × 10^3 J kg^−1 K^−1)(9 °C) Q_water = 11340 J
Next, let's calculate the heat lost by the tungsten and gained by the water. We can use the formula:
Q = mcΔT
Given: m_tungsten = 50 g = 0.05 kg c_tungsten = 1.05 × 10^2 J kg^−1 K^−1 ΔT_tungsten = T_flame - 22 °C
Using the formula, we have: Q_tungsten = (0.05 kg)(1.05 × 10^2 J kg^−1 K^−1)(T_flame - 22 °C)
Since the heat lost by the tungsten is equal to the heat gained by the water, we can set up the equation:
Q_tungsten = Q_water
(0.05 kg)(1.05 × 10^2 J kg^−1 K^−1)(T_flame - 22 °C) = 11340 J
Now we can solve for T_flame:
(0.05 kg)(1.05 × 10^2 J kg^−1 K^−1)(T_flame - 22 °C) = 11340 J
Simplifying the equation:
(0.0525 × 10^2 J K^−1)(T_flame - 22 °C) = 11340 J
(5.25 J K^−1)(T_flame - 22 °C) = 11340 J
Dividing both sides by (5.25 J K^−1):
T_flame - 22 °C = 2160 K
Adding 22 °C to both sides:
T_flame = 2182 °C
Therefore, the temperature of the flame is 2182 °C.
Similar Questions
A piece of tungsten of mass 50 g is placed over aflame for long time. The metal is then quicklytransferred to a well-insulated aluminumcalorimeter of mass 120 g containing 300 g ofwater at 22 °C. After some time the temperatureof the water reaches a maximum value of 31 °C.You may use these values of specific heatcapacity : water 4.2 × 10 J kg K ;tungsten 1.05 × 10 J kg K ;aluminium 9.45 × 10 J kg K . What is thetemperature of the flame? (Approx
A student uses a 350g block of tungsten to heat a 600ml beaker containing 450cm3 of tapwater. She adds exactly 2.5kJ of energy to the block before carefully placing it into the water.If the tap water was originally at 23oC, what would be the highest temperature that waterreaches? (assume that no heat is lost to the surroundings)
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).
A 33.0g sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C−1, is dropped into an insulated container containing 250.0g of water at 25.0°C and a constant pressure of 1atm. The initial temperature of the aluminum is 83.2°C.Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct number of significant digits.°C
A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The aluminum sample starts off at 90.6°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 32.3°C. The pressure remains constant at 1atm.Calculate the mass of the aluminum sample. Be sure your answer is rounded to the correct number of significant digits.
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.