What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy?
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
So, in order to find the final temperature of the water sample, you need to first find the change in temperature,
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.