Respuesta :
Explanation:
The given data is as follows.
mass, m = 75 g
[tex]T_{1} = 0^{o}C[/tex]
[tex]T_{2} = 27^{o}C[/tex]
Specific heat of water = 4.18
First, we will calculate the heat required for water is as follows.
q = [tex]m C \times (T_{1} - T_{2})[/tex]
= [tex]75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C[/tex]
= 8464.5 J/mol
= 8.46 kJ ......... (1)
Also, it is given that [tex]T_{3} = -20^{o}C[/tex] = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.
Now, we will calculate the heat of fusion as follows.
q = [tex]mC \times (T_{3} - T_{1})[/tex]
= [tex]0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K[/tex]
= -46.32 kJ ......... (2)
Now, adding both equations (1) and (2) as follows.
8.46 kJ - 46.32 kJ
= -37.86 kJ
Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at [tex]27.0^{o}C[/tex] to ice at [tex]-20.0^{o}C[/tex] is -37.86 kJ.
Total 36.53 kJ energy is released in changing 75g of water at 27°C to ice at -20°C.
To change 75g of water at 27°C to ice at -20°C, we would require to remove the heat energy from the water.
Ther are following procesees involved:
(i) cooling of water 27°C to 0°C
the amount of energy released is ΔQ = mcΔT
where m is the mass of water = 75g
c is the specific heat capacity of water = 4.184 J/g·K
and ΔT = 0-27 = -27°C change in temperature
⇒ ΔQ = 75 × 4.184 × (-27)
⇒ ΔQ = -8472.6 J
8472.6 j energy is released
(ii) conversion of water to ice
ΔQ = mL
where L is the latent heat of fusion
ΔQ = 75 × 333
ΔQ = 24975 J energy is released
(iii) cooling of ice from 0°C to -20°C
ΔQ = mcΔT
here c = specific heat capacity of ice = 2.06 J/g
ΔT= -20-0 = -20°C change in temperature
⇒ ΔQ = 75 × 2.06 × (-20)
⇒ ΔQ = -3090 J
3090 J enerfy is released.
Adding the energies released in processes (i), (ii), and (iii) we get
Q = 8472.6 + 24975 + 3090
Q = 36537.6 J
Total 36.53 kJ energy is released.
Learn moe about specific heat capacity:
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