Question

The specific heat of liquid bromine is 0.226J/g-K and the density is 3.12g/mL, how much heat (J) is required to raise the temperature of 10.0mL of bromine from 25.00 C to 27.30 C?

Answer 1

`"16.2 J"`

Explanation:

The idea here is that you need to use bromine's density to determine how many grams of bromine you have in that `"10.0-mL"` sample.

A density of `"3.12 g/mL"` tells you that you get a mass of `"3.12 g"` for every `"100 mL"` of bromine. This means that your sample will have a mass of

`10.0color(red)(cancel(color(black)("mL"))) * "3.12 g"/(1color(red)(cancel(color(black)("mL")))) = "31.2 g"`

Now, the equation that establishes a relationship between heat absorbed and change in temperature looks like this

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

`q` - heat absorbed
`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

Plug in your values and solve for `q` to get

`q = 31.2color(red)(cancel(color(black)("g"))) * 0.226"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.30 - 25.00)color(red)(cancel(color(black)(""^@"C")))`

`q = color(green)("16.2 J")`