Question

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.

Answer 1

`c_(e1)` = 128.3 J / kg ° C

Step-by-step explanation:

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

Q = m
`c_(e)` 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
`c_(e1)` (T₁ -
`T_(f)`) = (m₂
`c_(e2)` + C) (
`T_(f)` - T₂)

`c_(e1)`= (m₂ ce2 + C) (
`T_(f)`-T₀) / (m₁ (T₁-
`T_(f)`)

`c_(e1)` = (85.45 10-3 4186 + 1.56) (25.63 - 23.82) / (69.12 10-3 (98.93 - 25.63))

`c_(e1)` = (357.69 + 1.56) 1.81 / (69.12 10-3 73.3)

`c_(e1)` = 650.24 / 5.0665

`c_(e1)` = 128.3 J / kg ° C