Question

Pre Lab for Calorimetry Introduction Calorimetry is used to determine the specific heats and heat of transformation of materials. One can carry this out in one of two ways. One way is to take a material and isolate it from the environment and then put it a know power (energy per unit time) and measure how the temperature changes with time. When the material undergoes a phase transition (such as from ice to water) the temperature will stay constant until the transition is complete allowing the heat of transformation to be measured, When the material is in a single phase the temperature will increase with the rate of increase depending on the heat capacity. The other way one can use calorimetry is to take two materials, one with a know heat capacity and one that is unknown and bring them into thermal contact. Once they have reached equilibrium one can calculate the heat capacity of the unknown based on the final temperature In this lab you will utilize both methods to measure the heat capacity of various objects. As practice answer the following questions: Consider a block of ice with a mass of 458 g at -30 C that is isolated from the environment. Heat is added at the rate of 100J/s. 1) How long does it take the ice to warm to 0 C 2) How long does it take the ice to melt at 0 C 3) How long does it take the water at 0 C to warm to 100C 4) Take a large pot that can hold at least eight quarts and put it on your stove. Fill it with 2 quarts of water and measure the temperature with a kitchen thermometer. (If you don't have a thermometer you can estimate the initial temperature). Turn on the burner to high and measure the time it takes to start boiling. What is the power that your stove burner can deliver? 5) Repeat the experiment in 4 with 6 quarts of water. Does it take exactly six times as long? Explain?

Answer 1

Solution:  specific heat of water= 4.18 J/gc

specific heat of ice=2.03 J/gc

heat addition rate=100 J/s

1)

heat required for ice at -30 to reach 0(Q)= ms$\Delta$T

m=mass of ice=458g

s= specific heat of ice=2.03 J/gc

$\Delta$T=change in tempreture=30

Q=458*2.03*30=27892.2J

time taken to supply this amount of heat= Q/ rate of heat addition

= 27892.2/100

=278.92 seconds

2)  heat required to melt the ice(Q)= ml

m = mass of ice=458 g

L= latent heat of melting=334 J/g

Q=458* 334=1529272 J

time required for melting of ice= Q/rate of heat addition

=152972/100

=1529.72 seconds

3) heat required to raise tempreture from 0 to 100(Q)= ms$\Delta$T

Q=458*4.18.100

time taken to raise this tempreture= Q/rate of heat addition

=458*4.18*100/100

=1914.44 seconds

4) heat required for boiling=mL

m=mass of water=458g

L= latent heat of vaporization=2260J/g

time taken to boil the water=heat required/ rate of heat addition

= 458*2260/100=10350.8 seconds

power of stove= heat addition rate

5) time will become 6 times because mass will become 6 times so heat required.