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.
