In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 69.12 grams of tungsten to 98.93 °C and then drops it into a cup containing 85.45 grams of water at 23.82 °C. She measures the final temperature to be 25.63 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.56 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of tungsten.

Respuesta :

Answer:

 [tex]c_{e1}[/tex] = 128.3 J / kg ° C

Explanation:

In this exercise we will use that the expression for heat is

    Q = m [tex]c_{e}[/tex] ΔT

As they indicate that there are no losses with the medium, the heat transferred by the tungsten is equal to the heat absorbed by the water plus the calorimeter

    Q assigned = QAbsorbed

    Q hot = Q cold + Q calorimeter

The mass of tungsten (m₁ = 69.12 10⁻³ kg) with an initial temperature (T₁ = 98.93°C),

The mass of water (m₂ = 85.45 10⁻³ kg) at a temperature (T₂ = 23.82°C),

a calorimeter constant (C = 1.56 J/ °C)

m₁ [tex]c_{e1}[/tex] (T₁ - [tex]T_{f}[/tex]) = (m₂ [tex]c_{e2}[/tex] + C) ([tex]T_{f}[/tex] - T₂)  

[tex]c_{e1}[/tex]= (m₂ ce2 + C) ([tex]T_{f}[/tex]-T₀) / (m₁ (T₁-[tex]T_{f}[/tex])  

[tex]c_{e1}[/tex] = (85.45 10-3 4186 + 1.56) (25.63 - 23.82) / (69.12 10-3 (98.93 - 25.63))  

[tex]c_{e1}[/tex] = (357.69 + 1.56) 1.81 / (69.12 10-3 73.3)  

[tex]c_{e1}[/tex] = 650.24 / 5.0665  

[tex]c_{e1}[/tex] = 128.3 J / kg ° C