# Question #779cf

##### 1 Answer

#### Explanation:

In order to be able to solve this problem, you must know the **specific heat** of water, which you'll find listed as

#c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)#

Now, the specific heat of a given substance tells you the amount of heat needed to increase the temperature of

In the case of water, you have

#c_"water" = color(blue)("4.18 J") color(white)(.)color(red)("g"^(-1) color(white)(.)color(purple)(""^@"C"^(-1))#

which means that you need

Now, you need to raise the temperature of

#DeltaT = 80.0^@"C" - 25.0^@"C" = 55.0^@"C"#

The first thing to do here is to calculate how much heat will be needed to raise the temperature of *conversion factor*

#85.0 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "355.3 J" ""^@"C"^(-1)#

This means that **for every**

In your case, you will need

#55.0 color(red)(cancel(color(black)(""^@"C"))) * "355.3 J"/(1color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("19,500 J")))#

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

You can double-check this result by using the equation

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

Here

#q# is the heat lost or gained by the substance#m# is themassof the sample#c# is thespecific heatof the substance#DeltaT# is thechange in temperature, defined as the difference between thefinal temperatureand theinitial temperatureof the sample

In your case, you will have

#q = 85.0 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 55.0 color(red)(cancel(color(black)(""^@"C")))#

#q = color(darkgreen)(ul(color(black)("19,500 J")))#