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

1 Answer
Nov 17, 2015

Answer:

#66^@"C"#

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 #4.18"J"/("g" ""^@"C")#.

A substance's specific heat tells you how much heat is required to increase the temperature of #"1.0 g"# of that substance by #1^@"C"#.

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

#q# - the amount of heat
#m# - the mass of the sample
#c# - its specific heat
#DeltaT# - the change in temperature, defined s 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, #DeltaT#. Since heat was absorbed by the water, you must have a positive value for #DeltaT#.

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.