# What was the heat lost from the system if #"5.25 g"# of water was cooled from #64.8^@ "C"# to #5.5^@ "C"#? #C_P = "4.184 J/g"^@ "C"#

##### 1 Answer

#### Explanation:

Set up the following processes to solve the problem with ease:

#m=5.25g# , the mass of water

#C_p=(4.186J)/(g*^oC)# , specific heat capacity of#H_2O_(liquid)# ; http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html

#T_f=5.5^oC# , the initial temperature

#T_i=64.8^oC# , the final temperature

#Q_"lost"=?# This is a negative value , as heat moves out from the system.

The heat flow is found based on the mass involved, its heat capacity, and the changes of temperature in the process; thus,

#Q=mC_pDeltaT# where:

#DeltaT=T_f-T_i# Now, plug in identified value to each variable in the formula as shown below. Make sure the units work out and the desired unit as required is attained :

#Q=mC_p(T_f-T_i)#

#Q=5.25cancel(g)xx(4.186J)/cancel(g*^@C)xx(5.5-64.8)cancel(*^@C)#

#Q=color(red)(-)ul ?J# or

#Q_"lost"=ul?J#

Note:

The negative sign

#color(red)(-)# here indicates heat lost in the process.