How much heat (in Joules) is gained by the water when a chemical reaction takes place in 100.0 ml aqueous solution, has a temperature increase of 12.67°C and a specific heat capacity of 4.180 J/°Cg? [Hint: ignore any effects of the calorimeter and solve for qwater]
H (joules)
=ΔT*m*Q
where
H=energy required (joules)
ΔT=change in temperature (°C)
m=mass (g)
Q=specific heat in j/°C/g
So
heat gain
=12.67 °C * 100 g * 4.180 j/°C/g
=5296 j
Why did the water go to the party? Because it heard there would be a hot chemical reaction! Now, let's calculate the heat gained by the water using the given information.
To find the heat gained by the water, we can use the equation:
q = mcΔT
where:
q is the heat gained by the water,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
Given that the volume of the aqueous solution is 100.0 mL, we can assume the density of water to be 1 g/mL, so the mass of the water is 100.0 g.
Now, substituting the values into the equation:
q = (100.0 g)(4.180 J/°Cg)(12.67°C)
To solve this equation, I'll let my calculator do the math...
Calculating...
The heat gained by the water during the chemical reaction is approximately equal to 52,737.86 Joules.
Keep in mind that this calculation does not include any effects of the calorimeter. But hey, at least we didn't ignore the water's temperature increase—water hates being ignored!
To determine the amount of heat gained by the water (qwater), we can use the formula:
qwater = mass × specific heat capacity × temperature change
First, we need to find the mass of the water. Since we are given the volume of the solution (100.0 ml), we can assume that the density of water is approximately 1 g/ml. Therefore, the mass of the water is:
mass = volume × density
mass = 100.0 ml × 1 g/ml
mass = 100.0 g
Next, we can substitute the values into the formula to calculate qwater:
qwater = 100.0 g × 4.180 J/°Cg × 12.67°C
qwater = 52813.6 J
Therefore, the water gains 52,813.6 Joules of heat during the chemical reaction.
To find the amount of heat gained by the water, we need to use the formula:
q = m × c × ΔT
where:
q is the amount of heat gained or lost (in Joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (in J/°Cg)
ΔT is the change in temperature (in °C)
In this case, we are given the following information:
Volume of the solution (V) = 100.0 ml
Temperature increase (ΔT) = 12.67°C
Specific heat capacity of water (c) = 4.180 J/°Cg
First, we need to find the mass of the water. We can do this by converting the volume of the solution to grams, assuming the density of water is 1 g/ml.
Density of water = 1 g/ml
Mass of water (m) = Volume of solution × Density of water
= 100.0 ml × 1 g/ml
= 100.0 g
Now, we can substitute the values into the formula:
q = m × c × ΔT
= 100.0 g × 4.180 J/°Cg × 12.67°C
= 5326.36 J
Therefore, the amount of heat gained by the water is 5326.36 Joules.