# Question #54a28

##### 1 Answer

#### Answer:

#### Explanation:

The equation that you need to use here looks like this

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

Here

#q# is the heatlost or gainedby the water#m# is themassof water#c# is thespecific heatof water, equal to#"4.184 J g"^(-1)""^@"C"^(-1)# #DeltaT# is thechange in temperature

Now, you know that your sample of water has a mass of *joules per gram Celsius* tells you that you need to convert the mass of water from

*kilograms*to

*grams*.

#95 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 95 * 10^3color(white)(.)"g"#

You also know that the temperature of the water must *increase* by **change in temperature** is equal to

This basically means that you need to assume that the water starts at a given temperature **as liquid water** and ends at a temperature that is **as liquid water**, i.e. that you're not dealing with a *phase change* here.

So all you have to do now is to simply plug in your values into the equation and find the amount of heat *needed* for your sample.

#q = 95 * 10^3 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 30color(red)(cancel(color(black)(""^@"C")))#

#q = color(darkgreen)(ul(color(black)("12,000,000 J")))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you have one significant figure for the change in temperature.