From law of conservation of energy,
heat lost by unknown solid = heat gained by water
m1*s1*DT1 = m2*s2*DT2
m1 = mass of unknown solid = 42.5 g
s1 = specific heat of glass = x j/g.c
DT1 = (200-26.40) c
m2 = mass of water = 100 g
s2 = specific heat of water = 4.184 j/g.c
DT2 = (26.40-20) c
therefore, 42.5*x*(200-26.40) = 100*4.184*(26.40-20)
x = 0.3629 j/g.c
Therefore, specific heat of the unknown solid = 0.363 j/g.c
4 .184J/( "C))intay at 20.0 °C. Aher the A mass of 42.5g of an unknown solid...
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