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.

  = 128.3 J / kg ° C

Explanation:

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

    Q = m 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 (T₁ - ) = (m₂ + C) ( - T₂)  

= (m₂ ce2 + C) (-T₀) / (m₁ (T₁-)  

= (85.45 10-3 4186 + 1.56) (25.63 - 23.82) / (69.12 10-3 (98.93 - 25.63))  

= (357.69 + 1.56) 1.81 / (69.12 10-3 73.3)  

= 650.24 / 5.0665  

= 128.3 J / kg ° C

okay i have so many questions. but the big one here is: um wt..f is this mess? ?

come closer and become far apart and become steam/gas