A student calibrates his/her calorimeter. S/he performs part 1 as instructed. The data are: 99.2 g water at 36.0°C are added to 97.3 g water in the calorimeter at 16.0°C. The final temperature is 26.0°C. Calculate the heat capacity of the calorimeter in J/°C.
To calculate the heat capacity of the calorimeter, we need to use the principle of energy conservation.
The heat gained by the cold water is equal to the heat lost by the hot water plus the heat absorbed by the calorimeter. The heat gained or lost by a substance can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, the heat gained by the cold water is equal to the heat lost by the hot water:
mcΔT (cold water) = mcΔT (hot water) + CΔT (calorimeter)
Now let's plug in the given values:
m (cold water) = 99.2 g
c (water) = 4.18 J/g°C (specific heat capacity of water)
ΔT (cold water) = final temperature - initial temperature = 26.0°C - 16.0°C = 10.0°C
m (hot water) = 97.3 g
ΔT (hot water) = final temperature - initial temperature = 26.0°C - 36.0°C = -10.0°C (negative since the temperature decreased)
Now we can calculate the heat capacities of the water and the calorimeter:
Q (cold water) = mcΔT = (99.2 g)(4.18 J/g°C)(10.0°C)
Q (hot water) = mcΔT = (97.3 g)(4.18 J/g°C)(-10.0°C)
Now let's rearrange the equation to solve for the heat capacity of the calorimeter (CΔT):
CΔT (calorimeter) = Q (cold water) - Q (hot water)
CΔT (calorimeter) = (99.2 g)(4.18 J/g°C)(10.0°C) - (97.3 g)(4.18 J/g°C)(-10.0°C)
Now calculate this to get the value of CΔT (calorimeter).