Answer:
a. Energy lost, Q = 16,800 Joules.
b. Power = 28 J/s
c. Time, t = 2357.14 seconds
d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
Explanation:
Given the following data;
- Mass = 0.20 kg
- Initial temperature, T1 = 20°C
- Final temperature = 0°C
- Time = 10 minutes
a. To find the energy lost by the water as it cools to 0 degree celsius;
Mathematically, heat capacity is given by the formula;
[tex] Q = mcdt [/tex]
Where;
- Q represents the heat capacity or quantity of heat.
- M represents the mass of an object.
- C represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 20 - 0
dt = 20°C
We know that the specific heat capacity of water is equal to 4200 J/kg°C
Substituting the values into the formula, we have;
[tex] Q = 0.20 * 4200 * 20 [/tex]
Energy lost, Q = 16,800 Joules.
b. To find the average rate at which the water is losing energy in J/s by using the following formula;
[tex] Power = \frac {energy}{time} [/tex]
First of all, we would have to convert the value of time in minutes to seconds.
Conversion:
1 minute = 60 seconds
10 minutes = X seconds
Cross-multiplying, we have;
X = 60 * 10
X = 600 seconds
Substituting the values into the formula, we have;
[tex] Power = \frac {16800}{600} [/tex]
Power = 28 J/s
c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;
We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.
Mathematically, the latent heat of fusion is calculated by using the formula;
Energy, Q = ml = pt
Substituting the values into the formula, we have;
0.20 * 3.3 * 10⁵ = 28 * t
0.20 * 330000 = 28t
66000 = 28t
[tex] t = \frac {66000}{28} [/tex]
Time, t = 2357.14 seconds.
d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.
