Q: A kilogram ( kg ) of water with a temperature of 20o C is poured into a calorimeter. Subsequently, an unknown mass of stainless steel with a temperature of 80o C is placed in the water. What must the mass of the steel be in order for the temperature of the water to rise to the midpoint temperature of 50o C?
A: As is always the case, heat energy ( Q ) will move from a region of high temperature to a lower one. The exact amount of energy that is transferred will not only depend upon the molecular characteristics of the mediums in question; the quantity of the masses involved will also determine how the thermodynamic transaction occurs.
– Q lost = + Q gained
Q = mcΔT
ΔT = Tfinal – Tinitial
– mscsΔTs = mwcwΔTw
ms = ( mwcwΔTw / – csΔTs )
ms = [ mwcw( 50o C – 20o C ) / – cs( 50o C – 80o C ) ]
ms = [ mwcw( 30o C ) / – cs( – 30o C ) ]
ms = [ mwcw( 30o C ) / cs( 30o C ) ]
ms = ( mwcw / cs )
ms = ( 1 kg )( cw / cs )
Note: Since the temperature increments on the Kelvin and Celcius scales are the same size, we may substitute ( K ) or ( C ) accordingly when temperature differences must be taken into account.
ms = ( 1 kg )[ ( 4,181 J / kg*K ) / ( 490 J / kg*K ) ]
ms = ( 1 kg )( 8.53 )
ms = 8.53 kg
Wow!!!
This result, however, does indeed make physical sense. Since it takes 8.53 times as much heat energy to change the temperature of a kilogram of water by 1 Kelvin, it will take 8.53 times as much mass of steel to deliver the heat energy that establishes thermal equilibrium at the midpoint temperature in question. The temperatures of the steel and water may have changed by the same amount, but the quantities of steel and water needed to bring about this change must differ in accordance with their ability to accommodate heat energy. Within nature, water plays a role in maintaining a stable internal energy within many living things, and it plays a key role in determining the flow and behavior of ocean currents and overall atmospheric weather conditions as well.
RENAISSANCE PHYSICS 