Question

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?

Answer 1

`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"))`