In an experiment
Volume cold water
to determine
the heat of solution
of a substance,
the following
information was obtained.
of Volume of
Temperature
Temperature of hot water
Specific heat
of Temperature of water
hot water
of cold water
capacity
water
mixture
10 ml
10 ml
25 °C
35°C
4.2 cal/g/°C
28 °C
Use the information to answer questions 32, 33, and 34,
32. Heat lost by warm water is (A) 420 cal (8) 410 cal (C) 400 cal (D) 240 cal
33. Heat gained by cooler water is (A) 126 ca) (8) 200 cal (C) 120 cal (D) 220 cal 34. Water equivalent of the flask is (A) 98 cal/°C (B) 100 cal/°C (C) 128 cal/°C (D) 96 cal/°C
To answer these questions, we need to calculate the heat lost and gained by the water using the formula:
Q = mcΔT
Where Q is the heat gained/lost, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
For Question 32:
Heat lost by warm water = Heat gained by cooler water
Heat gained by cooler water = Heat lost by warm water = 10 ml * 4.2 cal/g/°C * (35°C - 28°C) = 294 cal
Therefore, the heat lost by the warm water is 294 cal, which is closest to (A) 420 cal.
For Question 33:
Heat gained by cooler water = 294 cal
Therefore, the heat gained by the cooler water is 294 cal, which is closest to (A) 126 cal.
For Question 34:
Water equivalent of the flask = (Heat lost by warm water - Heat gained by cooler water) / (Temperature of water mixture - Temperature of hot water)
Water equivalent of the flask = (294 cal - 294 cal) / (28°C - 35°C)
Water equivalent of the flask = 0 cal / -7°C
Therefore, the water equivalent of the flask is 0 cal/°C, which is closest to (D) 96 cal/°C.