CH 202 Lab 1 Report Name: Data and Calculations for Coffee-Cup Calorimetry Run 1

Question

CH 202 Lab 1 Report Name: Data and Calculations for Coffee-Cup Calorimetry Run 1 Run 2 1. Unknown Number 2. Mass of metal 3. Mass of Styrofoam cup 4. Mass of water and Styrofoam cup 5. Mass of water 6. Initial temperature of water in cup from graph C 33.1-C 7. Temperature of boiling water/metal shot 8. Final temperature of water + metal shot from graph a 3.5-u.C 9. Atwater Atsurr 10·Atmetal-Atsystem 11. Energy gained by water (use specific heat of water-4.1841/g-o and qsm .urr 100 looc 12. Specific heat of unknown metal (remember qys -Qsur) 13. Average value of specific heat Page 5

Explanation / Answer

Ans. TRIAL 1:

Step 1: Heat absorbed by water during increase in temperature:

q = m s dT                            - equation 1

Where,

q = heat required

m = mass of solution

s = specific heat of water = 4.184 J g-10C-1

dT = Final temperature – Initial temperature = Tf - Ti = (23.50C -22.7) = 0.80C

# Putting the values in above equation-

            q = 100.0 g x (4.184 J g-10C-1) x 0.80C = 334.72 J

Therefore,

Total amount of heat gained by water = 334.72 J

# Step 2: The total amount of “heat gained by water during its temperature increase from 22.70C to 23.50C” must be equal to the amount of “heat lost by metal sample while it cools from 100.00C to 23.50C”.

So,

            Total heat lost by metal = -334.72 J

(Note: The -ve sign indicates loss of heat).

# Let the specific heat of metal sample be s.

Putting the values for metal sample in equation 1-

            -334.72 J = 30.249 g x s x (23.5 – 100.0)0C

            Or, s = 334.72 J / (2314.0485 g 0C)

            Hence, s = 0.145 J g-10C-1

Therefore, specific heat of metal = s = 0.145 J g-10C-1

#TRIAL 2:

Step 1: dT = Tf - Ti = (24.00C -23.7) = 0.30C

# Putting the values in above equation-

            q = 100.0 g x (4.184 J g-10C-1) x 0.30C = 125.52 J

Therefore,

Total amount of heat gained by water = 125.52 J

# Step 2: The total amount of “heat gained by water during its temperature increase must be equal to the amount of “heat lost by metal sample while it.

So,

            Total heat lost by metal = -125.52 J

# Let the specific heat of metal sample be s.

Putting the values for metal sample in equation 1-

            -125.52 J = 30.249 g x s x (24.0 – 100.0)0C

            Or, s = 125.52 J / (2298.924 g 0C)

            Hence, s = 0.055 J g-10C-1

Therefore, specific heat of metal = s = 0.055 J g-10C-1

# Average specific heat = [s (trial 1) + s (trial 2)] / 2

                                                = (0.145 J g-10C-1 + 0.055 J g-10C-1) / 2

                                                = 0.100 J g-10C-1

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