Question #e69fc

1 Answer
Jun 14, 2016

Answer:

#117.39g#, rounded to two decimal places

Explanation:

We know that in such interactions heat is gained by one and is lost is by the other.

Also that the heat gained/lost is given by
#DeltaQ=mst# ....(1)
where #m,s and t# are the mass, specific heat and rise or gain in temperature of the object;
#Delta Q_"lost"=Delta Q_"gained"# .....(2)

In the given problem heat is lost by sample of copper and gained by water.

Heat gained by water to change its temperature from #21.0^@"C"# to #29.7^@"C"#
#DeltaQ_"gained"=mst#, Specific heat of water is #1calg^-1 "^@C^-1#.
#DeltaQ_"gained"=298xx1xx(29.7-21.0)#
#=>DeltaQ_"gained"=2592.6cal#

Heat lost by the sample to change its temperature from #275.1.^@"C"# to #29.7^@"C"#, specific heat of copper#=0.09calg^-1 "^@C^-1#. If #m_s# is the mass of sample
#Delta Q_"lost"=m_sxx0.09xx(275.1-29.7)#
#=>Delta Q_"lost"=22.086xxm_s cal#

Using (2) we get

#22.086xxm_s=2592.6#
#=>m_s=117.39g#, rounded to two decimal places