] Week 6 Calorimetry 163L Calorimetry pdf CHEM 163L Spring 2018- vas iastate.edu
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] Week 6 Calorimetry 163L Calorimetry pdf CHEM 163L Spring 2018- vas iastate.edu/courses/32560/quizzes/120343/take (specific heat for water is 4.18J/gC). Do not enter unit M Week 6 Calorimetry. Invit ts in your answer or it will be counted wrong. Make sure you use correct significant figures. 12.540 Question 3 15 pts A 10.000g ice cube is added to 100.000g of water at 80.0°C in a calorimeter. The final temperature of the water in the calorimeter is 65.5 °C. How much heat in kilojoules was needed to melt the ice cube? (specie heat for water is 4.18 J/g C) Do not include units in your answer or it will be counted wrong. Use correct significant figures. Question 4 5 pts Match each of the following errors with the effect they would have on calculated heat (g) compared to actual heat. Answers may be used more than once The initial temperature for hot water was measured a few minutes before mixing and then the water was allowed to sit at room temperature I Choose ] DOLLExplanation / Answer
Q = mcT
Q = heat energy (Joules, J), m = mass of a substance (kg)
c = specific heat (units J/kgK), is a symbol meaning "the change in"
T = change in temperature (Kelvins, K)
The heat gained by ice = The heat lost by water.
Heat of fusion = 334.16 J g¯1
specific heat capacity for solid water (ice) = 2.06 J g¯1 K¯1
specific heat capacity for liquid water = 4.184 J g¯1 K¯1
10 x 10 x 2.06 + 10 x 334.16 + 10 x 4.18 x 65.5 = 36359.9 Joules
The total Heat was needed to bring the ice to 65.5 Deg Cel = 36359.9 Joules = 36.3599 Kilo Joules
Heat was needed to melt the ice = 10 x 10 x 2.06 + 10 x 334.16 = 33622 Joules = 33.622 Kilo Joules
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