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Sample Problem p.633: Heat transferred from water to ice
In the ThoughtLab, 10.0 g of ice was added to 60.0 g of water. The initial temperature of the water was 26.5°C.
The final temperature of the mixture was 9.7°C. How much heat was lost by the water?
Known:
Mass of water (m) = 60.0 g
c (water- liquid) = 4.184 J/g·°C (see photocopied table)
Initial temperature (Ti) = 26.5°C (c)
Final temperature (Tf) = 9.7°C
Note: Because you are concerned only with the water, do not include the mass of the ice in the calculations.
Solution:
i) ΔT = Tf − Ti = 9.7°C - 26.5°C = -16.8 °C
ii) q = mcΔT = (60.0 g)(4.184 J/g·°C )(-16.8°C) = −4217.472 J
iii) Round your answer to the appropriate number of significant digits
(3 digits = least number of digits in the multiplication procedure q = mcΔT values known)
= −4220 J or −4.22 × 103 J or − 4.22 kJ (Note: 1000 J = 1kJ)
Check Your Solution: The water lost heat, so the heat value should be negative. Heat is measured in joules or
kilojoules. Make sure that the units cancel out to give the appropriate unit for your answer.
Sample Problem p. 635: Calculating Specific Heat Capacity
Calculate the specific heat capacity of canola oil, using the information below from Part B of the ThoughtLab
on page 631. Note that the ice gained 4.20 × 103 J of energy when it came in contact with the canola oil.
Known: Mass of oil (m) = 60.0 g
q system (oil) = - q surroundings (ice) = - 4.0 × 103 J
Solution:
q = mcΔT
→
q =c
mΔT
→
Initial temperature of oil (Ti) = 35.0°C
Final temperature of oil (Tf) = 5.2°C
−4.0 × 103 J
=c
(60.0 g)(5.2°C − 35.0°C)
−4.0 × 103 J
=c
(60.0 g)(29.8°C)
−4.0 × 103 J
=c
-1788 g°C
= 2.237136465 J / g·°C (round to the appropriate number of sig figs)
= 2.24 J/ g·°C
Check Your Solution: The specific heat capacity should be positive, and it is. It should have the units J / g·°C.
Please complete questions #1=10