Question

Scoring Scheme: 3-3-2-1 Part III. For each trial, enter the amount of heat gained by the chemical system of ammonium nitrate, qrxn. The specific heat of water is 4.184 J/goC. Report your answer using 4 digits. Trial

Answer 1

Question:

The information given is:

Trial # Tiwater T f ΔT Masswater (m)

#1: 21.2 10.8 10.8 24.990

#2: 20.8 9.50 9.5 25.000

#3: 20.9 9.20 9.2 25.010

Answer:

The heat of the reaction is -5985 J

Step-by-step explanation:

The heat absorbed by the water is given by

ΔQ = m·c·ΔT

From which

∑ (ΔT·m)/3 = 278.34 kg·°C

ΔQ = c×∑ (ΔT·m)/3 = 4.184 J/g·°C×278.34 kg·°C = 1164.565 J

ΔQ Calorimter = Specific heat capacity of calorimeter,
`c_(calorimeter)` × ΔT
`_(average)`

Where the
`c_(calorimeter)` = 443 J/°C for example, we have

ΔQ Calorimter = 443×11.133 = 4820.733 J

From which the heat of reaction is then

`\Delta Q_(reaction) = -(\Delta Q_(water) + \Delta Q_(calorimter))`

`\Delta Q_(reaction) = -5985.298 \, J`

Using 4 digits, we get

`\Delta Q_(reaction) \approx -5985 \, J`.