The process of accurately determining the specific heat capacity ( c ) of a substance is dependent upon minimal heat ( Q ) losses occurring within a system’s frame of reference. A calorimeter is designed so that such losses are negligible to the extent that they can be ignored. Consider the diagram below:
A thin metal cup with a low mass and specific heat is surrounded with an insulated outer layer. The top is tightly sealed so that temperature measurements and stirring of the medium can be done remotely with convenience. Somewhat analogous to this system are food items that are wrapped in tin foil prior to being placed in an oven.
Q: A calorimeter with a thin inner surface made of 150 g of copper contains 500 g of water at 20.0 ℃. An unidentified 225-g object is heated to 508 ℃ and subsequently submerged into the water medium. Within a few minutes, the water maintains a temperature of 40.0 ℃. What is the specific heat capacity of the sample?
A: In previous examples, we determined the specific heat capacity and specific heat of objects in separate steps to compare and contrast them. In many scenarios, this need not be the case. In this example, we will carry out all necessary multiplication steps at once. Prior to doing so, we develop a general accounting scheme to conceptually track the flow of heat energy from the unknown material to the water medium:
– Qs = Qc + Qw
As is always the case, the quantity of heat lost by the unknown substance is gained by the water and the metal cup. Fortunately, the mathematics works in such a way that each term is positive prior to final calculations being made:
– mscsΔTs = mcccΔTc + mwcwΔTw
– ( 0.225 kg )( cs )( 40.0 ℃ – 508 ℃ ) = ( 0.150 kg )( 390 J / kg*K )( 20 K ) + ( 0.500 kg )( 4,186 J /kg*K )( 20 K )
Note: Once again, we must ensure that the proper units are being used throughout the equation. The temperature differential of the unknown object has been recorded with Celsius units. Fortunately, temperature differences on the Celsius scale are directly proportional to those of the Kelvin scale. With this being the case, we substitute the Celsius values with Kelvin units:
– ( 0.225 kg )( cs )( 40.0 K – 508 K ) = ( 0.150 kg )( 390 J / kg*K )( 20 K ) + ( 0.500 kg )( 4,186 J /kg*K )( 20 K )
– ( 0.225 kg )( cs )( – 468 K ) = ( 0.150 kg )( 390 J / kg*K )( 20 K ) + ( 0.500 kg )( 4,186 J /kg*K )( 20 K )
( 105.3 kg*K )( cs ) = ( 1,170 J + 41,860 J )
( 105.3 kg*K )( cs ) = 43,030 J
cs = 409 J / kg*K
RENAISSANCE PHYSICS 