Question

A 0.52g sample of metal that looks like gold requires 1.65 J of energy to change its temperature from 25.0°C to 40°C. Is this metal pure gold?

Answer 1

No, it's not.

Explanation:

The idea here is that you need to use the information provided by the problem to calculate the specific heat of the unknown material.

A quick comparison with the listed value for the specific heat of gold will determine if the unknown material is indeed pure gold or not.

So, your tool of choice here will be the equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat lost or gained by the substance
• `m` is the mass of the sample
• `c` is the specific heat of the substance
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

Rearrange to solve for `c`

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

Plug in your values to find

`c = "1.65 J"/("0.52 g" * (40.0 - 25.0)^@"C") = "0.212 J g"^(-1)""^@"C"^(-1)`

Now, gold has a specific heat of about

`c_"gold" = "0.129 J g"^(-1)""^@"C"^(-1)`

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

which means that your unknown material is not pure gold. The fact that its specific heat is higher than that of pure gold implies that it also contains materials that have a higher specific heat than that of gold.