What minimum number of 1.0 kg stones at 530 ∘C must be added to a vessel holding 6.0 kg of 20∘C water to bring the water to a boil? Use 800 J/(kg⋅K) for the specific heat of the stones and 4190 J/(kg⋅K) for the specific heat of water.

**Answer 1**

Let m be the number of kilos of stone to be dropped. The boiling temperature is 100 degrees. Now we equate the energies. Water needs energy

6*4190*(100-20)=2011200 J.

Stone is also at 100 degrees. It has energy. m *100*800=m=80000 J. This must ebergy the stone should provide.

m* 530* 800= 2011200 + m 80000. This gives

m(424000-.80000)= 2011200.

m 344000=2011200,

m=5,85 kg.