50mL of water at 46.9 degrees Celsius were mixed with 50mL of water at 25.2 degrees Celsius in a calorimeter also at 25.1 degrees Celsius. the final temp was 30.1 Celsius. assuming that neither the desity of water nor its specific heat capacity changed with temperature, calculate the total heat capacity of the calorimeter. [desity of water = 1.00 g/ml and specific heat capacity = 4.18kJ/gK]

Question

50mL of water at 46.9 degrees Celsius were mixed with 50mL of water at 25.2 degrees Celsius in a calorimeter also at 25.1 degrees Celsius. the final temp was 30.1 Celsius. assuming that neither the desity of water nor its specific heat capacity changed with temperature, calculate the total heat capacity of the calorimeter. [desity of water = 1.00 g/ml and specific heat capacity = 4.18kJ/gK]

*if someone has already amswered this question, could you provide a link to the answer please ! :) thanks.*

Answer 1

I think this is the one.

http://www.jiskha.com/display.cgi?id=1352834709

Answer 2

thank you :)

Answer 3

To calculate the total heat capacity of the calorimeter, we can use the concept of heat transfer.

The heat lost by the 50mL of water at 46.9 degrees Celsius (initial temperature) can be calculated using the formula:

Q1 = m1 * c1 * (Tf - Ti1)

Where:
Q1 is the heat lost by the hot water
m1 is the mass of the hot water
c1 is the specific heat capacity of water
Tf is the final temperature
Ti1 is the initial temperature of the hot water

Substituting the values into the formula:
Q1 = 50g * 4.18 kJ/gK * (30.1°C - 46.9°C)

Similarly, the heat gained by the 50mL of water at 25.2 degrees Celsius (initial temperature) can be calculated using the formula:

Q2 = m2 * c2 * (Tf - Ti2)

Where:
Q2 is the heat gained by the cold water
m2 is the mass of the cold water
c2 is the specific heat capacity of water
Tf is the final temperature
Ti2 is the initial temperature of the cold water

Substituting the values into the formula:
Q2 = 50g * 4.18 kJ/gK * (30.1°C - 25.2°C)

Now, since the total heat lost by the hot water is equal to the total heat gained by the cold water, we have:

Q1 = Q2

50g * 4.18 kJ/gK * (30.1°C - 46.9°C) = 50g * 4.18 kJ/gK * (30.1°C - 25.2°C)

To find the total heat capacity of the calorimeter, we can rearrange the equation as:

Ccal * (Tf - Ti) = Q1

Where:
Ccal is the total heat capacity of the calorimeter
Ti is the initial temperature of the calorimeter

Substituting the values into the equation:
Ccal * (30.1°C - 25.1°C) = 50g * 4.18 kJ/gK * (30.1°C - 46.9°C)

Now, we can solve for Ccal:

Ccal = (50g * 4.18 kJ/gK * (30.1°C - 46.9°C)) / (30.1°C - 25.1°C)

Calculating this expression will give you the total heat capacity of the calorimeter.