# Question #c1015

##### 1 Answer

#### Answer:

#### Explanation:

In order to solve this problem, you need to know water's *specific heat*, which expresses the amount of heat needed to raise the temperature of

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

Now, the amount of heat is given to you in *calories*, which means that you're going to have to convert it to *Joules*

#1000color(red)(cancel(color(black)("cal"))) * "4.18400 J"/(1color(red)(cancel(color(black)("cal")))) = "4184.0 J"#

The equation that establishes a relationship between heat added/removed and increase/decrease in temperature looks like this

#q = m * c * DeltaT" "# , where

*final temperature* and the *initial temperature*.

Rearrange the equation to solve for

#DeltaT = q/(m * c)#

#DeltaT = (4184.0color(red)(cancel(color(black)("J"))))/(100color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ^@"C")) = 10.0 ""^@"C"#

This means that the final temperature of the water will be

#DeltaT = T_"final" - T_"initial"#

#T_"final" = "DeltaT" + T_"initial" = 10""^@"C" + 40""^@"C" = color(green)(50""^@"C")#