Question

The quantity of heat Q that changes the temperature L1Tof a mass mof a substance isgiven by Q = cmt:T, where c is the specific heat capacity of the substance. Forexample,forH20,c=1caljg'C",Andfora change of phase, the quantity of heat Q that changes the phase of a mass m is Q = ml., where L is the heat of fusion or heat of vaporization of the substance. For example, for H20, the heat offusion is 80 cal/g (or 80 kcaljkg) and the heat of vaporization is 540 cal/g (or 540 kcaljkg). Use these relationships to determine the number of calories to change (a) 1 kg ofO°C ice to O°C ice water, (b) 1 kg ofO°C ice water to 1 kg of 100°C boiling water, (c) 1 kg of 100°C boiling water to 1 kg of 100°C steam, and (d) 1 kg ofO°C ice to 1 kg of 100°C steam.

Answer 1

Answer:

a) Q = 80,000 cal

b) Q = 100,000 cal

c) Q = 540,000 cal

d) Q = 720,000 cal

Explanation:

a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion

$Q_{l} = ML_{f}$ = 1 * 80

$Q_{l}$ = 80 kCal = 80,000 cal

b) 1 kg of O°C ice water to 1 kg of 100°C boiling water

Specific heat capacity, c =  1000cal/kg.C

$Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal$

c) 1 kg of 100°C boiling water to 1 kg of 100°C steam

Latent heat of vaporization is needed for this conversion

$Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal$

d)  1 kg of O°C ice to 1 kg of 100°C steam.

Q = $Q_{L} + Q_{c} + Q_{v}$

Q = 80,000 + 100,000 + 540,000

Q = 720,000 cal