Since no change in the state of matter took place (water remained liquid while the
brass cylinder remained solid), use the formula ΔQ = m c ΔT = m c [ T – T ₀ ] for
each body that absorbed or gave off heat … where … c = specific heat of the material
… T₀ = initial temperature … T = final temperature … since heat flows from high to
low temperature, the body which is initially at a higher temperature (the brass
cylinder) lost heat, while the body that is initially at the lower temperature (water)
gained heat … actually, heat must also flow into the calorimeter cup containing the
water … but since no information regarding the calorimeter cup was given, we just
assume that it is made of thermally insulating material, so that all the heat given off
was absorbed by the water … for the brass cylinder (1), we have …
… ΔQ₁ = m₁ c₁ ΔT₁ = ( 410 g ) c₁ [ 32.0°C – 95.0°C ] = ( 410 g ) c₁ [ - 63.0°C ] …
since the brass cylinder must also have the same final temperature as that of the
heated water… that is … ΔQ₁ = - c₁ ( 25,830 ) g • °C …
For water (2) … ΔQ₂ = m₂ c₂ ΔT₂ = ( 335 g ) [ 4.18 J / ( g • °C ) ] [ 32.0°C – 25.0°C ]
… ΔQ₂ = ( 335 g ) [ 4.18 J / ( g • °C ) ] [ 7.0°C ] = 9802.1 J … by conservation of
energy … ΔQ₁ + ΔQ₂ = 0 … --> … [ c₁ ( - 25,830 ) g • °C ] + 9802.1 J = 0 … solving
for c₁ , we finally get … c₁ = 9802.1 J / [ ( - 25,830 ) g • °C ] = 0.380 J / ( g • °C )
