Question

The specific heat of a solution is 4.18 J/(g•°C)and its density is 1.02 g/mL. The solution is formed by combining 25.0 mL of solution A with 25.0 mL of solution B with each solution initially at 21.4°C. The final temperature of the combined solution is 25.3°C. Calculate the heat of reaction, q, assuming no heat loss due to the calorimeter.

I got 831 J
The part I cannot figure out is the question afterwards which is If the calorimeter has a heat capacity of 8.20 J/°C and a correction is included to account for the heat absorbed by the calorimeter what is the heat of reaction q

Answer 1

Okay, so, to solve for this, we're going to have to use q = mcΔT. However, the mass of the calorimeter is not important because not of it is used in the reaction, so really, we are only looking at 2 things, the temperature change and specific heat. So, here is the lightly modified equation we will use:
q = cΔT

Now, just plug in the ΔT we had for the original equation (which was 3.9) and use the specific heat of the calorimeter to get q.

q = (8.20) * (3.9)
q = 31.98, or about 32

The 32 is what was absorbed by the calorimeter, so we can add that to the original value to get our answer.
831 + 32 = 863 J

So, the answer is 863 J

Hope this helped!! :D

Answer 2

a. qrxn = 831 J

b. 863 J

Step-by-step explanation:

we know that density is the mass of a substance per unit volume

d=mass/volume

the volume of the solution is the combination of solution A and solution B

1.02 g/mL=mass/(25+25)

mass=50*1.02

mass=51g

Recall that Q=mCdT

mass=m, C=specific heat capacity

dT=change in temperature

qrxn = (51 g)(4.18 J/g⋅°C)(25.3 °C - 21.4 °C)

qrxn = 831 J

2.Heat=Heat capacity *change in temperature

qcal = (8.20 J/°C)((25.3 °C - 21.4 °C)

qcal = 31.98 J

qrxni + qcal = qrxn

qrxn = 831 J + 32.0 J

863 J ------Heat of reaction

863 J =(51 g)(Heat Capacity)(25.3 °C - 21.4 °C)

4.34 J/g⋅°C