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)= msT
m=mass of ice=458g
s= specific heat of ice=2.03 J/gc
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)=
msT
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.
Pre Lab for Calorimetry Introduction Calorimetry is used to determine the specific heats and heat of...
The second pic is the experiment result.
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Coffee Cup Calorimetry: Specific Heat
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