Can you help me solve these post lab questions (with steps)? Thank you! CALCULAT

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Can you help me solve these post lab questions (with steps)? Thank you!

CALCULATIONS For a close system, which is approximated by a calorimeter, the heat gained and lost by each part must add to zero, so: 1. qcal + qwater + qmetal = 0 2. Calculate qcal from the equation: q 1 = Ccal where 7-(Tr al-Ti itial) for the wa 3. Calculate qwater from the equation qwater-mwater C T' where mwater is the mass of the 4. Combining these three equations and rearranging allows you to calculate qmetal from 5. The value calculated for qmetal For the heat capacity of the calorimeter, use the value Coal = 2001 K-1 water in the calorimeter, Cwater 4.184 J g1 K-1, and AT (Tinal - Tinitial) for the water. qmetal-mwater C AT cold -Ccal ATcold. specific heat capacity of the metal using the equation qmetal mmetal Cmetal AT metal using the equation in step 5 can be used to determine the metal is the mass of the metal used in your experiment and AT metal Tfinal-100.0 C We assume that the metal pieces start at the temperature of boiling water, or 100.0°0C From your three data trials, calculate three values of Cmetal and a using the actual value of Cmetal you obtained from your instructor, calculate the percent verage them. Then, error in the average value of Cmetal you determined. SAMPLE DATA/RESULTS TABLE (NOT FOR RECORDING YOUR DATA) Trial 1 Trial 2 Trial 3 Data Mass of water in the calorimeter (g) Mass of metal pieces added to calorimeter (g) Initial temp, water plus calorimeter (oC) Initial temp, metal pieces(C) Final temp (°C AT for water + calorimeter (K) T for metal pieces added to calorimeter (K) 4.237534.0 IS-01 | 23.214 , 21 7.421 | 26.808 100.000 1975 32.136 31.68 4.76' 5.007 y.8H 100.000 100.000 Trial 1 Trial 2 Trial 3 Results q for the water in the calorimeter ) q for the calorimeter (U), using Ccal 20.0JK1 q for the metal (U) Heat capacity of the metal (Ug K) Average heat capacity of the metal (U g K Accepted value of the heat capacity of the metal (Jg K % error in the heat capacity of the metal (J g-1

Explanation / Answer

For trial 1:

dT for the metal pieces added to the calorimeter = (31.975-100) = -68.0250C

q for water in the calorimeter = mwater*C*dTwater = 39.42*4.184*4.761 = 785.25 J

q for the calorimeter = Ccal*dTcal = 20*4.761 = 95.22 J

Using the equation given, q for the metal = -(qwater + qcal) = -(785.25+95.22) = -880.47 J

Heat capacity of the metal = qmetal/(mmetal * dTmetal)

Putting values we get:

cmetal = -880.47/(15.01*(-68.025)) = 0.862 J/(g.K)

In this way calculate the values for trials 2 and 3 and then obtain the average value.

Hope this helps !

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