Question

A tray containing 0.20kg of water at 20degree celsius is placed in a freezer. The temperature of the water drops to 0degree celsius in 10 minutes. Calculate
a) The energy lost by the water asit cools to 0 degree celsius.

b) The average rate at which the water is losing energy in J/s.

c) Estimate the time taken for the water at 0 degree celsius to turn completely into ice.

d) state any assumptios you make​

Answer 1

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;

$Q = mcdt$

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;

$Q = 0.20 * 4200 * 20$

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;

$Power = \frac {energy}{time}$

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;

$Power = \frac {16800}{600}$

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

$t = \frac {66000}{28}$

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.

Answer 2

Nice, I think is really good