Question

The initial temperature of the water in a constant-pressure calorimeter is

14°C. A reaction takes place in the calorimeter, and the temperature rises
to 87°C. The calorimeter contains 254 g of water, which has a specific heat
of 4.18 J/(g.°C). Calculate the enthalpy change during this reaction. *

Answer 1

Answer: The enthalpy change during this reaction is 77505.56 J.

Step-by-step explanation:

Given:
`T_(1) = 14^(o)C`,
`T_(2) = 87^(o)C`

Mass = 254 g, Specific heat =
`4.18 J/g^(o)C`

Formula used to calculate the enthalpy change is as follows.

`q = m * C * (T_(2) - T_(1))`

where,

q = enthalpy change

m = mass of substance

C = specific heat capacity

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\= 254 g * 4.18 J/g^(o)C * (87 - 14)^(o)C\\= 77505.56 J`

Thus, we can conclude that the enthalpy change during this reaction is 77505.56 J.