HEAT AND THERMAL ENERGY: Specific Heat Capacity of an Unknown Metal

Q: A sample of an unknown metal is heated to a predetermined temperature. It is subsequently placed within a quantity of water that is room temperature. The temperature of the water is determined to have risen by a certain amount after being measured a second time around. Generally speaking, how may this information be used to determine the specific heat capacity ( c ) of the metal?

A: In order to understand the concept of specific heat capacity, it is fundamental that students thoroughly understand the difference between heat and temperature. Heat is a measure of vibrational energy within a material, whereas temperature can be regarded as a concentration of heat energy within a given mass of matter. With this being said, two equal quantities of different materials cannot be assumed to contain the same quantity of heat if they are the same temperature!!! This same rule applies to different masses of different materials that are the same temperature, even though the quantity of heat energy held at a given temperature is indeed mass-dependent.

Consider, for example, an oven that has been heated to 350^o F. If a glass dish within the oven contains food that is loosely wrapped in aluminum foil, the foil, glass, and food will be in thermal equilibrium with one another; however, it takes a lot less heat energy to establish a 350^o F temperature within the foil. If the foil and food are carefully removed from the hot oven, the glass dish will be hot to touch for far longer than the foil in question. This is due to differences in a material’s capacity to accommodate and manage heat energy. The formal definition of specific heat capacity ( c ) is that it’s the amount of heat energy ( J ) needed to raise a given mass ( m ) of a substance by 1 Kelvin:

c = Joules per unit mass per Kelvin.

c = ( J / m ) / K

c = ( J / mK )

Whether mass is expressed in grams ( g ) or kilograms ( kg ) depends upon the subject being covered; however, the temperature in question is always in regard to the Kelvin scale. Since the Kelvin and Celcius scales are divided into same-sized increments, the differences in temperature on a Celsius scale are directly proportional to differences in temperature on a Kelvin scale:

ΔT_C ∝ ΔT_K

And furthermore,

ΔT = ( T_final – T_initial )

Taking all of the aforementioned information into account, the quantity of heat ( Q ) absorbed or liberated by a substance with a particular specific heat capacity is mathematically expressed in the following manner:

Q = mcΔT

Let’s now revisit the original question at hand. If a heated sample of metal is placed within a sample quantity of water at room temperature, an exchange of heat energy will occur; however, if it takes a proportionally greater quantity of heat energy to change to the temperature of water by 1 K, the metal will have a decrease in temperature of a much greater magnitude than the temperature increase of the water solution!!! This is in fact what is observed under such circumstances. Water has an unusually high specific heat capacity, and this makes it a useful solvent to use within a calorimeter. Summing things up, an amount of heat energy transferred from a metal to a quantity of water will cause less of a change of temperature to occur in the water, but the temperature of the water will increase until thermal equilibrium is established throughout the fluid medium:

– Q lost = Q gained

– m_mc_mΔT_m = m_wc_wΔT_w

c_m = ( m_wc_wΔT_w / – m_mΔT_m)

It may seem as if we’ve run into an unfortunate mathematical conundrum with the minus sign in the denominator; however, the metal underwent a decrease in temperature:

ΔT_m = ( T_{m final} – T_{m initial} )

T_{m final} < T_{m initial} 

( T_{m final} – T_{m initial} ) = – ΔT_m

c_m = [ m_wc_w( T_{w final} – T_{w initial} ) / – m_m( T_{m final} – T_{m initial} ) ]

c_m = ( m_wc_wΔT_w / – m_m -ΔT_m )

c_m = ( m_wc_wΔT_w / m_mΔT_m )