Thermochemistry Pre-Lab Assignment 11

Question

A 56.3-g sample of water was heated to 51.2 C and added to 45.8 g of 20.3 C water in a calorimeter. Assume that no heat was lost to the surroundings including the calorimeter. Assume that no heat was lost to the surrounding including the calorimeter, calculate the final temperature of the resulting solution. If the final temperature for Problem 3 were 34.7 C, how much heat was lost to the calorimeter? Calculate the heat capacity for the calorimeter

Answer 1

Heat will be transferred from the warmer object (51.2º water) to the cooler object (20.3º water) until an equilibrium temperature (final temperature, Tf) is reached. Obviously, the heat LOST by the warm water must be equal to the heat gained by the cool water. The equation that relates these variables is..

q = mC∆T

q = heat; m = mass; C = specific heat; ∆T = change in temperature

For the warmer water we have the following:

q = (56.3 g)(4.184 J/gº)(51.2º - Tf)

For the cooler water we have the following:

q = (45.8g)(4.184 J/gº)(Tf - 20.3º) note use Tf-20.3 so that ∆T value will be positive

Since heat lost by warm = heat gained by cool, we can write...

(56.3 g)(4.184 J/gº)(51.2º - Tf) = (45.8g)(4.184 J/gº)(Tf - 20.3º)

(56.3 g)(51.2º - Tf) = (45.8g)(Tf - 20.3º)

2882.56 - 56.3Tf = 45.8Tf - 929.74

102.1Tf = 3812.3

Tf = 37.3ºC

We don't know exactly what was involved in Problem 3, but if the conditions were the same as in this problem, then we can find the heat that was gained by the calorimeter, and hence we can calculate the heat capacity (C_cal) of the calorimeter.

The final temperature in Problem 3 is 34.7 instead of our calculated temperature of 37.3º. This tells us that some heat went into the calorimeter. We can find this amount of heat by comparing the heat lost from the warm water to that gained by the cooler water

heat lost from warm water = q = mC∆T = (56.3g)(4.184 J/gº)(51.2 - 34.7) = 3887 J

heat gained by cooler water = q = mC∆T = (45.8g)(4.184 J/gº)(34.7 - 20.3º) = 2759 J

heat gained by the calorimeter = 3887 J - 2759 J = 1128 J

For the calorimeter ... q_calorimeter = C_cal∆T

q_calorimeter = heat gained by calorimeter = 1128 J

C_cal = calorimeter constant = ?

∆T = T_final - T_{initial cool water} = 34.7º - 20.3º = 14.4º

Solving for C_cal we get C_cal = (q_calorimeter)/(∆T) = 1128 J / 14.4º

C_cal = 78.3 J/º

Answer 2

Q = mC(T_f - T_i)

For water C (the specific heat capacity) = 4.186 J/g°C

Let the 51.2 °C water be sample #1 and let the 20.3 °C water be sample #2.

Sample #1 gives off heat to warm the solution and Sample #2 accepts heat to warm the solution. The heat given off by Sample #1 is the same as the heat absorbed by sample #2 except Q₁ is a negative (since it is given off and Q2 is a positive (since it is being absorbed). So -Q₁ = Q₂

m₁ = 56.3 g and T_i = 51.2 °C so Q₁ = (56.3)(C)(Tf - 51.2)

m₂ = 45.8 g and T_i = 20.3 °C so Q₂ = (45.8)(C)(Tf - 20.3)

The samples are both water meaning the "C" values are equal so they cancel out and, of course, the final temp is the same for both:

-Q₁ = Q₂

-(56.3)(C)(T_f - 51.2) = (45.8)(C)(T_f - 20.3)

-(56.3)(T_f - 51.2) = (45.8)(T_f - 20.3)

-56.3Tf + 2882.56 = 45.8T_f - 929.74

3812.3 = 102.1T_f

T_f = 3812.3/102.1 = 37.3°C