A 410-g cylinder brass is heated to 95 C and placed in a calorimeter containing 335 g of water at 25 C. The water is stirred, and its highest temp is recorded as 32 C. from the thermal energy determine the specific heat of brass. The specific heat of water is 4.18 j/g C

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)