Dear Depthy,Please find enclosed a doc file with the data analysis, calculations and the answers to the different questions.Since the main objective is that you understand and learn how to solve this kind of problems on your own next time, I would advise you to go through them carefully and come back to me in case you have any doubts so that we can clarify them. Thus, I would gladly answer any questions you might have regarding this assignment.Kind regards,Carmen
DATA & CALCULATIONS
1. Show typed calculations for the following values: mhot, ∆Thot,
qhot, mcool, ∆Tcool, qcool, qcal, ∆Tcal, and Ccal.
2. Create a table of just the calculated values listed above, displaying each value
with proper units, mathematical sign and correct number of significant figures.
In order to solve part B you should consider that there is an equilibrium such that the heat of
hot water is transferred to cold water and the calorimeter walls.
Considering that heat transfer can be calculated as:
Q = mass * Specific heat * Variation of temperature
and the process involved, we can establish the equation:
Q (hot water) = Q (cold water) + Q (calorimeter)
m (HOT) * Ce (water) * ∆T (HOT) = m (COLD) * Ce (water) * ∆T (COLD) + m (cal) * Ce (cal) * ∆T (cal)
where as you see I simply substituted each heat term by the corresponding formula with its
According to the data of your laboratory results we know:
Hot water: 85.0 mL at 79 ºC
Cold water: 50.0 mL at 24.6 ºC
Final equilibrium temperature: 58.2 ºC
Calculation of the mass of water (valid for both mhot and mcool):
In order to calculate the mass we will apply the formula:
mass = density * Volume
where we can assume the density of both cold and hot water is equal to 1 since the
experimental error will be greater than the one resulting from not considering the density
correction for different temperatures.
Thus, we find:
m(hot) = 1 g/ml * 85.0 g/ml = 85.0 g
m (cold) = 1 g/ml * 50.0 ml = 50.0 g
Calculation of the variation of the temperature (valid for both ∆Thot...