Calculate the Calorimeter Constant (Ccal) if 25.00 g of water at 51.50°C was added to 25.00 g of water at 25.00 °C with a resulting temperature of 35.00 °C?

(*remember the specific heat capacity of water (cwater) = 4.180 J/g°C)

Hot water lost

Δ H₁=CmΔT₁ =4.180• 25•(51.5-35) =1724 J
Cold water got
Δ H₂=CmΔT₂ =4.180• 25•( 35-25) =1045 J
Δ H = Δ H₁-Δ H₂=1724-1045=679 J
heat capacity of the calorimeter (Calorimeter Constant)
C(cal)=Δ H/ΔT = 679/10 =67.9 J/℃

To calculate the calorimeter constant (Ccal), we need to use the principle of heat transfer or the equation for heat exchange, which is

q = m × c × ΔT

where:
q is the heat energy transferred (in this case, from the hotter water to the colder water),
m is the mass of the water (both the initial and final masses are the same),
c is the specific heat capacity of water, and
ΔT is the change in temperature of the water.

First, let's calculate the heat energy gained or lost by each water sample.

For the initial water at 51.50°C:
q1 = m × c × ΔT
q1 = 25.00 g × 4.180 J/g°C × (35.00°C - 51.50°C)

And for the initial water at 25.00°C:
q2 = m × c × ΔT
q2 = 25.00 g × 4.180 J/g°C × (35.00°C - 25.00°C)

Since heat energy is conserved, the heat gained by the colder water (q2) is equal to the heat lost by the hotter water (q1). Therefore,

q1 = -q2

Substituting the known values into the equations:

25.00 g × 4.180 J/g°C × (35.00°C - 51.50°C) = - (25.00 g × 4.180 J/g°C × (35.00°C - 25.00°C))

Next, we can solve for the calorimeter constant (Ccal) using the equation:

q = Ccal × ΔT

Since q is known to be equal to q1, we can rearrange the equation:

Ccal = q1 / ΔT

Plugging in the value for q1 we just calculated and the ΔT of the water samples:

Ccal = (25.00 g × 4.180 J/g°C × (35.00°C - 51.50°C)) / (35.00°C - 25.00°C)

Simplifying this expression will give you the value for the calorimeter constant (Ccal).