Question

How much heat must be added to a 8.0-kg block of ice at -8 C to change it to water at 14 C? The specific heat of ice is 2050 J/kg x C, the specific heat of water is 4186 J/kg x C, the latent heat of fusion of ice is 334,000 J/kg, and 1 cal = 4.186 J.

Answer 1

To change the 8.0-kg block of ice at -8°C to water at 14°C, we need to consider the heat required to raise the temperature of the ice from -8°C to 0°C, the heat required for phase change from ice to water at 0°C, and the heat required to raise the temperature of the water from 0°C to 14°C. The total heat required is 3,271,376 J.

Step-by-step explanation:

To change the 8.0-kg block of ice at -8°C to water at 14°C, we need to consider the heat required to raise the temperature of the ice from -8°C to 0°C, the heat required for phase change from ice to water at 0°C, and the heat required to raise the temperature of the water from 0°C to 14°C.

The heat required to raise the temperature of the ice is calculated using the formula:

Q = mass × specific heat × temperature change

Substituting the values, we get:

Q₁ = 8.0 kg × 2050 J/kg°C × (0°C - (-8°C)) = 131,200 J

The heat required for phase change is calculated using the formula:

Q = mass × latent heat

Substituting the values, we get:

Q₂ = 8.0 kg × 334,000 J/kg = 2,672,000 J

The heat required to raise the temperature of the water is calculated using the formula:

Q = mass × specific heat × temperature change

Substituting the values, we get:

Q₃ = 8.0 kg × 4186 J/kg°C × (14°C - 0°C) = 468,176 J

Finally, we can calculate the total heat required by summing up the individual heats:

Total heat = Q₁ + Q₂ + Q₃ = 131,200 J + 2,672,000 J + 468,176 J = 3,271,376 J

Answer 2

The total heat required is the sum of heat needed to raise the temperature of ice to 0 degrees Celsius, the heat needed for the phase change from ice to water, and the heat required to raise the temperature of the resulting water to 14 degrees Celsius.

Step-by-step explanation:

To calculate how much heat must be added to an 8.0-kg block of ice to change it to water at 14℃, we must consider two steps: heating the ice up to 0℃, and then melting and heating the resulting water up to 14℃. The amount of heat required for each step is calculated using different properties of ice and water.

First, we calculate the heat needed to raise the temperature of ice from -8℃ to 0℃. The equation for this is:

Q1 = m * c_ice * ΔT

Where Q1 is the heat added, m is the mass (8.0 kg), c_ice is the specific heat of ice (2050 J/kg℃), and ΔT is the temperature change (8℃).

After the ice is at 0℃, we need to calculate the heat required for the phase change from ice to water at 0℃ using the latent heat of fusion of ice. This is given by:

Q2 = m * Lf

Where Q2 is the heat added and Lf is the latent heat of fusion (334,000 J/kg).

Next, we heat the water from 0℃ to 14℃. The equation for this is:

Q3 = m * c_water * ΔT

Where Q3 is the heat added, m is the mass (8.0 kg), c_water is the specific heat of water (4186 J/kg℃), and ΔT is the temperature change (14℃).

The total heat added is the sum of Q1, Q2, and Q3. Thus:

Q_total = Q1 + Q2 + Q3