# Question #da3fc

##### 1 Answer

#### Answer:

#### Explanation:

You know that the **specific heat** of water is equal to

Now, you know that your sample has a mass of

#51.4 color(red)(cancel(color(black)("g"))) * "4.184 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "215.06 J"""^@"C"^(-1)#

This tells you that in order to increase the temperature of

#92.7^@"C" - 5.6^@"C" = 87.1^@"C"#

which means that you will need a total energy input of

#87.1 color(red)(cancel(color(black)(""^@"C"))) * overbrace("215.06 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 51.4 g of water")) = color(darkgreen)(ul(color(black)("18,800 J")))#

The answer is rounded to three **sig figs**.

**SIDE NOTE** *Don't be confused by the fact that you have two sig figs for the initial temperature of the water.*

*Notice that you're using this value in a subtraction*

#92.7^@"C" - 5.6^@"C" = 87.1^@"C"#

*The result of this operation must be rounded to one decimal place. In this case*,

*is rounded to one decimal place and has*.

**three sig figs**, which is what you used in the subsequent multiplication