A 40.0 g block of ice at -15 degrees C is dropped into a calorimeter (of negligible heat capacity) containing water at 15 degrees C. When equilibrium is reached, the final temperature is 8.0 degrees C. How much water did the calorimeter contain initially? Specific heat of ice is 2090 J/(kg*K) and the latent heat of fusion of water is 33.5*10^4 J/kg. ..... answer is 546 g. How do I work this problem out?
3 answers
What specific heat are you to use for liquid H2O?
specific heat for liquid water is 4190 c (J/kg*K)
q1 for raising T of ice from -15 to zero C is
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)= ? (and I would work this as a separate step since you have all of the numbers and put that number for q1 below )
q2 to melt ice is
q2 = mass ice x heat fusion = ? (Again, you have all of the numbers; I would work this as a separate step and put that number in for q2 below in the final equation)
q3 = heat added to melted ice at zero C to the final T is
q3 = mass melted ice x specific heat water x (Tfinal-Tinitial) which is another separate step, then put that number into for q3 in the final equation below.)
q4 is the release of heat from the H2O in the calorimeter is
q4 = mass H2O x specific heat H2O x (Tfinal-Tinitial). Put this one in below as the equation with the one unnown of mass H2O.
Put all of them together for a zero sum as q1(separate number from above) + q2(separate number from above) + q3(separate number from above) + q4(place the equation for q4 here) = 0
Solve for mass H2O.
Post your work if you get stuck. I did a quickie calculation and about 550 g seems to be about right.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)= ? (and I would work this as a separate step since you have all of the numbers and put that number for q1 below )
q2 to melt ice is
q2 = mass ice x heat fusion = ? (Again, you have all of the numbers; I would work this as a separate step and put that number in for q2 below in the final equation)
q3 = heat added to melted ice at zero C to the final T is
q3 = mass melted ice x specific heat water x (Tfinal-Tinitial) which is another separate step, then put that number into for q3 in the final equation below.)
q4 is the release of heat from the H2O in the calorimeter is
q4 = mass H2O x specific heat H2O x (Tfinal-Tinitial). Put this one in below as the equation with the one unnown of mass H2O.
Put all of them together for a zero sum as q1(separate number from above) + q2(separate number from above) + q3(separate number from above) + q4(place the equation for q4 here) = 0
Solve for mass H2O.
Post your work if you get stuck. I did a quickie calculation and about 550 g seems to be about right.
3 answers