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
Question
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
Solution
To solve this problem, we need to use the principle of conservation of energy, which states that the total energy in a closed system remains constant. In this case, the energy gained by the water and the calorimeter is equal to the energy lost by the tungsten.
First, let's calculate the energy gained by the water and the calorimeter:
The energy gained by the water is given by the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the water: m = 300 g = 0.3 kg c = 4.2 × 10^3 J/kg/K ΔT = 31°C - 22°C = 9 K
So, Q_water = 0.3 kg * 4.2 × 10^3 J/kg/K * 9 K = 11,340 J
For the calorimeter: m = 120 g = 0.12 kg c = 9.45 × 10^2 J/kg/K ΔT = 9 K
So, Q_calorimeter = 0.12 kg * 9.45 × 10^2 J/kg/K * 9 K = 1,018 J
The total energy gained is Q_total = Q_water + Q_calorimeter = 11,340 J + 1,018 J = 12,358 J
This is the energy lost by the tungsten, so we can use this to find the initial temperature of the tungsten (which is the temperature of the flame). The energy lost by the tungsten is given by the formula Q = mcΔT, but we need to rearrange this to find the initial temperature: T_initial = Q/(mc) + T_final
For the tungsten: m = 50 g = 0.05 kg c = 1.05 × 10^2 J/kg/K Q = 12,358 J T_final = 31°C = 304 K
So, T_initial = 12,358 J / (0.05 kg * 1.05 × 10^2 J/kg/K) + 304 K = 2,348 K
Therefore, the temperature of the flame is approximately 2,348 K.
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