If a 40.0 grams of metal heated to 100degreeC. When this was placed in a calorim

Question

If a 40.0 grams of metal heated to 100degreeC. When this was placed in a calorimeter (assume a perfect calorimeter where only the water absorbs he heat) containing 100 grams of water (specific heat = 1.00 cal/g deg C) the calorimeter temperature rose from 20 degree C to 25 degree C. How much heat did the calorimeter gain? How much heat did the metal lose What is the specific heat of the metal important equation Q = SH X mass X delta T, SH = Q/mass times delta T SH = specific heat, Q = heat SH for water is 1 cal/g degree heat gained by water = heat lost by metal

Explanation / Answer

a)
Q = m*C*delta T
= 100*1*(25-20)
= 500 cal
Answer: 500 cal

b)
heat lost by metal = heat gained by calorimeter
metal lost 500 cal

c)
SH = Q / (m*delta T)
= 500 / (40*(100-25))
= 500/(40*75)
=0.17 cal /g.oC

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