Question

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

Answer 1

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

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.

$q = m \times C \times (T_{2} - T_{1})$

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.

$q = m \times C \times (T_{2} - T_{1})\\= 254 g \times 4.18 J/g^{o}C \times (87 - 14)^{o}C\\= 77505.56 J$

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