s Waller in the beaker placed on the hot plate. Tum on water in the beaker up to
ID: 2039327 • Letter: S
Question
s Waller in the beaker placed on the hot plate. Tum on water in the beaker up to 100°C (boiling water). Measure and record (in Tables below) the masses (m) of each of t and brass), then heat them by record (in Tables below) the initial temperature T, of the object inside the beaker. 2. e three objects (aluminum, stee suspending from a string in a beaker of boiling water. Measure an CAUSION: Both the object and the th water without touching the side or bottom of the beaker. ermometer tip must be completely immersed in the boiling g of room temperature water (m,) in the plastic cup. Measure and record (in Tables below) the initial temperature T with thermometer #2 (0-50°C) inside the cup. quickly. Measure and record (in Tables below) the final temperature T Repeat steps 3 to 5 three times for each object and Table 1 Measured data for specific heat of alurinum cover ve one object from the boiling water by holding its string, quickly place it in the cup, and cover the calorimeter so that the mixture reaches the final equilibrium temperature Place the object back to boiling water to heat it again. Pour out the water in the cup, dry the cup by paper towel. It is ready for the next experiment. record the diata in the following Tables. aluminumm(g) (C)my (8) 100 100 22.4 Table 2 Measured data for specific heat of steel 100 2 6 .93100 Table 3 Measured data for specific heat of brass brass m(g) 100 100 84Explanation / Answer
from table 1
m = 20.46 g
T1 = 100 C
mw = 100 g
Tw = 22.4 C
T2 = (25.4 + 25.8 + 25.1)/3 = 25.43333333 C
hence
jeat capacity of aluminium = c
mc(T1 - T2) = mw*1(T2 - Tw)
hence
c = 100(25.4333 - 22.4)/20.46*(100 - 25.4333)
c = 0.198821371336 cal/g C
similiarly
for steel
mc(T1 - T2) = mw*1(T2 - Tw)
hence
c = 100((28.3 + 27.9 + 28.0)/3 - 22.5)/67.93*(100 - (28.3 + 27.9 + 28.0)/3)
c = 0.11392 cal/g C
cfor brass
mc(T1 - T2) = mw*1(T2 - Tw)
hence
c = 100((29.1 + 29.4 + 30.0)/3 - 22.5)/111*(100 - (29.1 + 29.4 + 30.0)/3)
c = 0.0894511532809 cal/g C
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