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
