A student determined the calorimeter constant of the calorimeter. The student added 50.00 mL of cold water to 50.00 mK of heated , distilled water in a styrofoam cup. The initial temp of the cold water was 21,00C and the hot water, 29.15C. The maximum temp pf the mixture was found to be 24.81C. Assume the density of water is 1.00g mL-1 and the specific heats is 4.184 J g-1 K-1
It's asking me to find the heat lost by hot water
heat lost by hot water
q = mass hot water x specific heat H2O x (Tfinal-Tinitial)
To find the heat lost by the hot water, we need to use the principle of energy conservation. The heat lost by the hot water will be equal to the heat gained by the cold water and the calorimeter.
First, let's calculate the mass of the hot water:
Mass of hot water = volume of hot water x density of water
= 50.00 mL x 1.00 g/mL
= 50.00 g
Next, let's calculate the heat gained by the cold water and the calorimeter. We will use the formula:
Q = m x c x ΔT
where:
Q = heat gained/lost (in joules)
m = mass (in grams)
c = specific heat capacity (in J g-1 K-1)
ΔT = change in temperature (in degrees Celsius)
For the cold water and the calorimeter, the change in temperature is the final temperature minus the initial temperature:
ΔT = final temperature - initial temperature
= 24.81°C - 21.00°C
= 3.81°C
Now, let's calculate the heat gained by the cold water and the calorimeter:
Q = (mass of cold water + mass of calorimeter) x specific heat capacity x ΔT
Since the volume of cold water added is 50.00 mL, its mass can be calculated using the density of water:
Mass of cold water = volume of cold water x density of water
= 50.00 mL x 1.00 g/mL
= 50.00 g
Given that the specific heat capacity of water is 4.184 J g-1 K-1, we can now calculate the heat gained by the cold water and the calorimeter:
Q = (50.00 g + 50.00 g) x 4.184 J g-1 K-1 x 3.81°C
= 100.00 g x 4.184 J g-1 K-1 x 3.81°C
= 1600.00 J
Therefore, the heat lost by the hot water is 1600.00 J.