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A 21
A 21.5 g sample of a metal is immersed in boiling water at 100.0º C and then transferred to a calorimeter containing 75.0 g of water at 22.5ºC . The final temperature of the water is 27.0ºC. What is the specific heat of the metal?
Expert Solution
Using the concept of specific heat capacities,
heat released = heat absorbed
q = m * s * ?T
here m = mass
s = specific heat capacity
s of water = 4.2 j / g °c
?T = change in temperature
qmetal = qwater
21.5 * s * ( 100 - 27) °c = 75 * 4.2 * ( 27 - 22.5 ) °c
s = (315 * 4.5 )/ (21.5 * 73)
s = 1417.5 / 1569.5
s = 0.90 j / (g °c)
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