# What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy?

##### 1 Answer

#### Explanation:

In order to be able to solve this problem, you need to know the value of water's specific heat, which is usually given to be

A substance's specific heat tells you how much heat is required to increase the temperature of

Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* minus *initial temperature*

So, in order to find the final temperature of the water sample, you need to first find the *change* in temperature, **absorbed** by the water, you **must** have a positive value for

Plug in your values into the above equation to get

#q = m * c * DeltaT implies DeltaT = q/(m * c)#

#DeltaT = (47880color(red)(cancel(color(black)("J"))))/(380color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 30.14^@"C"#

Since the change in temperature is defined as the difference between the final temperature and the initial temperature, you can say that

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

#T_"final" = 36^@"C" + 30.14^@"C" = color(green)(66^@"C")#

The answer is rounded to two sig figs, the number of sig figs you have for the initial temperature and the mass of the sample.