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?

1 Answer
Nov 14, 2015

Answer:

#"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")#