Question

can you please help me plug in the correct number for number 5 so number 6 can be completed. Im not confident that my numbers were plugged in correctly.

Results Complete the following Data Table: Part A Finding the Specific Heat of a Metal 10.440 5,405 55.135 Mass of coffee cup calorimeter and lid Mass of test tube, stopper, and metal 55.475 Mass of empty test tube + stopper Mass of calorimeter, water, and lid 220c Initial temperature of water Initial temperature of metal 100.0°C 3290 Stluer Spial pakles Show calculations below and follow significant figure rules. Final temperature of water and metal Color of metal Shape or description of metal A. Finding the Specific Heat of a Metal 3.47 Calculate the mass of the metal sample. 1. 84.M-5.433 2. Calculate the mass of the water in the calorimeter. 60075-1014A5 44,59 3. Calculate the AT for the water. 320C-220C 4. Calculate the AT for the metal. -(st-) 78oc 100°c Page 7 of 11 Heat Flow and Calorimetry

Answer 1

To do this more understandable, let's do it in two steps. First, the heat gained by the water can be calculated as (all value for WATER: mass, heat capacity and delta T):

$q=m\cdot c\cdot \Delta T=49.58g\cdot 4.184\frac{J}{g\cdot ^{o}C}\cdot 10^{o}C=2074.43J$

Since energy must be conserved, this is the same amount of heat that has been "given" by the metal. Since it is "given" heat, it has a negative sign:

$q=-m\cdot c\cdot \Delta T$

And from here, we can rearrange to obtain the heat capacity of the metal (please take into consideration that, since the final temperature of the whole system is 32°C, the final temperature of the metal is 32°C, which makes delta T = -68°C)

$c=-\frac{q}{m\cdot \Delta T}=-\frac{2074.43J}{33.47g\cdot (-68)^{o}C}=0.911\frac{J}{g\cdot ^{o}C}$

And for the approximation of the molar mass, we can use the Petit-Dulong method, which states that:

$m_{molar}=\frac{25\frac{J}{mol\cdot K}}{c}=\frac{25\frac{J}{mol\cdot K}}{0.911\frac{J}{g\cdot K}}=27.3\frac{g}{mol}$

I have changed the units for the heat capacity of the metal to K instead of °C. This does not change the number, since it comes from the delta T value, ∆°C = ∆K.