A sample of an unknown metal has a mass of 120.7 g. As the sample cools from 90.5 °C to 25.7 °C, it releases 7020 J of energy. What is the specific heat of the sample?

1 Answer
Dec 18, 2015

Answer:

#c = 0.898"J"/("g" ""^@"C")#

Explanation:

As you know, a substance's specific heat tells you how much heat must be absorbed or lost in order for #"1 g"# of that substance to experience a #1^@"C"# temperature change.

The equation that establishes a relationship between heat absorbed / lost and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat absorbed/lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

Now, it is important to realize that the value of #q# must come out negative for samples that experience a decrease in temperature.

From a thermodynamic point of view, heat lost always carries a negative sign, so you need to keep that in mind when plugging in your values.

Simply put, when #"7020 J"# of energy are released, the heat lost is written as

#q = -"7020 J"#

With this being said, plug in your values into the above equation and solve for #c#, the specific heat of the metal

#q = m * c * DeltaT implies c = q/(m * DeltaT)#

#c = (-"7020 J")/("120.7 g" * (25.7 - 90.5)^@"C")#

#c = (color(red)(cancel(color(black)(-)))"7020 J")/(color(red)(cancel(color(black)(-)))"7821.36 g" ""^@"C") = 0.8975 "J"/("g" ""^@"C")#

Rounded to three sig figs, the number of sig figs you have for the two temperatures of the metal, the answer will be

#c = color(green)(0.898"J"/("g" ""^@"C"))#