A massless calorimeter contains 0.120 kg of H2O that is initially solid ice at a
ID: 1701410 • Letter: A
Question
A massless calorimeter contains 0.120 kg of H2O that is initially solid ice at a temperature of
-80 °C.
(a) How much heat does it take to raise the temperature of the H2O up to its melting/freezing
temperature of 0.0 °C, while the H2O is still frozen?
(b) How much heat does it take to melt the H2O at the constant temperature of 0.0 °C?
(c) How much heat does it take to heat the liquid H2O (water) up to its boiling point of 100.0 °C?
(d) How much heat does it take to boil off all the water at the constant temperature of 100.0 °C?
[The H2O vapor (steam) is allowed to escape from the calorimeter.]
(e) Assume that the heater in the calorimeter puts out heat at a constant rate of 20.0 Watt. Make a
table of the time in minutes after heat is first applied when the H2O first reaches 0.0 °C, when
the H2O is completely melted, when the H2O first reaches 100.0 °C, and finally when the H2O
is completely boiled off.
(f) Using a power point, plot the temperature of the H2O versus time in minutes from the time t = 0 at which the H2O is first heated to the time the H2O is all boiled off.
Explanation / Answer
Q = heat required to change the temperature of ice from -80.c to 0.c = mCt
where m = mass of ice = 0.120 Kg,
C= specific heat = 2100 J /kg C
t = change in temperature= 80
Q = heat required = (0.120)(2100)(80)=20160 joules
b) heat required to melt the ice without change in temperature= mL
m=mass = 0.120 Kg, L = latent heat of fusion; 3.34 X 105 J/kg
Q=(0.120)(3.34x 105)=40080 Joules
c)Heat required to heat the water from 0.C to 100.C = mCt
c= specific heat of water 4200 J /kgC
Q = (0.120)(4200)(100-0)
Q=50400 J
d)Heat required to convert water at 100.C to steam at 100.C = mL
L= latent heat of vaporization= 2268 x 103J/Kg
Q= (0.120)(2268 x 103)=272160 J
e)
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