A 0.20 kg ice cube at 0.0° C has sufficient heat added to it to cause total melting, and the resulting water is heated to70deg.Cel. > How much heat is added? (For water Lf=334kj/kg > andLv=2257kj/kg >

Well, well, well! It seems like we have an icy situation here. Let's break it down and melt away those calculations.

First things first, let's figure out how much heat is required to melt the ice cube. Since the ice starts at zero degrees Celsius, we need to use the latent heat of fusion (Lf) for ice. Lf is given as 334 kJ/kg.

The mass of the ice cube is 0.20 kg, so the heat required to melt the ice can be calculated using the formula: heat = mass * Lf

Heat = (0.20 kg) * (334 kJ/kg)
Heat = 66.8 kJ

Now, the ice has melted, and we have water at 0 degrees Celsius. We need to further heat this water to 70 degrees Celsius. But hold on, we have another parameter to consider here. The latent heat of vaporization (Lv) for water, which is 2257 kJ/kg.

Since we know the mass of the water is still 0.20 kg, we can calculate the heat required using the formula: heat = mass * Lv

Heat = (0.20 kg) * (2257 kJ/kg)
Heat = 451.4 kJ

So, my friend, adding the heat required to melt the ice cube (66.8 kJ) and the heat required to raise the resulting water to 70 degrees Celsius (451.4 kJ), we get a grand total of 518.2 kJ of heat added.

I hope I melted away any confusion you had with my chilling explanation!

To calculate the amount of heat added, we need to consider two stages: melting the ice cube and heating the resulting water.

Stage 1: Melting the ice cube
The heat required to melt the ice cube is calculated using the formula:

Q1 = m * Lf

Where:
Q1 is the heat required to melt the ice cube,
m is the mass of the ice cube, and
Lf is the latent heat of fusion of ice (334 kJ/kg).

Given:
m = 0.20 kg
Lf = 334 kJ/kg

Substituting the values into the formula:
Q1 = 0.20 kg * 334 kJ/kg
Q1 = 66.8 kJ

So, the amount of heat required to melt the ice cube is 66.8 kJ.

Stage 2: Heating the resulting water
The heat required to heat the water from 0°C to 70°C is calculated using the formula:

Q2 = m * Cp * ΔT

Where:
Q2 is the heat required to heat the water,
m is the mass of the water (which was originally the mass of the ice cube),
Cp is the specific heat capacity of water (4.18 kJ/kg°C), and
ΔT is the change in temperature (70°C - 0°C = 70°C).

Given:
m = 0.20 kg
Cp = 4.18 kJ/kg°C
ΔT = 70°C

Substituting the values into the formula:
Q2 = 0.20 kg * 4.18 kJ/kg°C * 70°C
Q2 = 58.52 kJ

So, the amount of heat required to heat the resulting water to 70°C is 58.52 kJ.

To find the total amount of heat added, we add Q1 and Q2:

Total heat added = Q1 + Q2
Total heat added = 66.8 kJ + 58.52 kJ
Total heat added = 125.32 kJ

Therefore, the total amount of heat added is 125.32 kJ.

To find the amount of heat added to melt the ice cube and heat the resulting water, we need to calculate the heat required for each step separately and then add them together.

Step 1: Heat required to melt the ice cube from 0.0°C to 0.0°C:

Since the ice is already at 0.0°C, it is in a solid form. To melt it, we use the formula:

Q1 = m * Lf

where Q1 is the heat required, m is the mass of the ice cube, and Lf is the latent heat of fusion for water.

Given that the mass of the ice cube is 0.20 kg and the latent heat of fusion for water, Lf, is 334 kJ/kg, we can calculate Q1:

Q1 = 0.20 kg * 334 kJ/kg = 66.8 kJ

So, the heat required to melt the ice cube is 66.8 kJ.

Step 2: Heat required to heat the resulting water from 0.0°C to 70°C:

Once the ice has melted, we need to heat the resulting water from 0.0°C to 70°C. The formula to calculate the heat required is:

Q2 = m * Cp * ΔT

where Q2 is the heat required, m is the mass of the water, Cp is the specific heat capacity of water, and ΔT is the change in temperature.

Given that the mass of the water is still 0.20 kg, the specific heat capacity of water, Cp, is approximately 4.18 kJ/(kg·°C), and the change in temperature, ΔT, is 70°C - 0.0°C = 70°C, we can calculate Q2:

Q2 = 0.20 kg * 4.18 kJ/(kg·°C) * 70°C = 58.52 kJ

So, the heat required to heat the resulting water to 70°C is 58.52 kJ.

Step 3: Total heat added:

To find the total heat added, we add the heat required for each step:

Total heat added = Q1 + Q2 = 66.8 kJ + 58.52 kJ = 125.32 kJ

Therefore, the total heat added is 125.32 kJ.

heat added: Lf*mass + mass*specificheat*(70C-0C)