how much heat is needed to convert 966g of ice at -10 deg C to steam at 126 deg C

To calculate the heat needed to convert ice at -10°C to steam at 126°C, we need to consider the three steps involved in the process:

1. Phase change from ice (-10°C) to water (0°C)
2. Heating water from 0°C to boiling point of water (100°C)
3. Phase change from water to steam (100°C to 126°C)

Let's break down and calculate each step:

Step 1: Phase change from ice to water
To calculate the heat required for this phase change, we use the formula:
Q = m × ΔHf

Where:
Q is the heat energy
m is the mass of the substance
ΔHf is the heat of fusion

The heat of fusion for ice is 334 J/g.

m = 966 g (given)
ΔHf = 334 J/g

Q1 = m × ΔHf
= 966 g × 334 J/g
= 322,044 J

Step 2: Heating water from 0°C to boiling point of water (100°C)
To calculate the heat required for this step, we use the formula:
Q = m × Cp × ΔT

Where:
Q is the heat energy
m is the mass of the substance
Cp is the specific heat capacity
ΔT is the change in temperature

The specific heat capacity of water is 4.18 J/g°C.

m = 966 g (given)
Cp = 4.18 J/g°C
ΔT = (100°C - 0°C) = 100°C

Q2 = m × Cp × ΔT
= 966 g × 4.18 J/g°C × 100°C
= 402,228 J

Step 3: Phase change from water to steam
To calculate the heat required for this phase change, we use the formula:
Q = m × ΔHv

Where:
Q is the heat energy
m is the mass of the substance
ΔHv is the heat of vaporization

The heat of vaporization for water is 2260 J/g.

m = 966 g (given)
ΔHv = 2260 J/g

Q3 = m × ΔHv
= 966 g × 2260 J/g
= 2,179,360 J

Now, we can add up the heat energy for each step to find the total heat needed:

Total heat = Q1 + Q2 + Q3
= 322,044 J + 402,228 J + 2,179,360 J
= 2,903,632 J

Therefore, around 2,903,632 J of heat energy is needed to convert 966g of ice at -10°C to steam at 126°C.

To calculate the amount of heat needed to convert ice at -10°C to steam at 126°C, we need to consider several steps of the phase change process.

The first step is to calculate the heat required to raise the temperature of the ice from -10°C to its melting point at 0°C. We can use the specific heat capacity of ice to calculate this. The specific heat capacity of ice is 2.09 J/g°C.

Heat required to raise the ice temperature from -10°C to 0°C:
Q1 = mass × specific heat capacity × temperature change
Q1 = 966 g × 2.09 J/g°C × (0°C - (-10°C))
Q1 = 966 g × 2.09 J/g°C × 10°C
Q1 = 20266.8 J

Next, we calculate the heat required to convert the ice at 0°C to liquid water at 0°C. This is known as the latent heat of fusion. The latent heat of fusion for water is 334 J/g.

Heat required for fusion of ice:
Q2 = mass × latent heat of fusion
Q2 = 966 g × 334 J/g
Q2 = 322644 J

After this, we calculate the heat required to raise the temperature of the liquid water from 0°C to 100°C. The specific heat capacity of water is 4.18 J/g°C.

Heat required to raise the temperature of water from 0°C to 100°C:
Q3 = mass × specific heat capacity × temperature change
Q3 = 966 g × 4.18 J/g°C × (100°C - 0°C)
Q3 = 403925.2 J

Next, we calculate the heat required for the phase change from water at 100°C to steam at 100°C. This is known as the latent heat of vaporization. The latent heat of vaporization for water is 2260 J/g.

Heat required for vaporization of water:
Q4 = mass × latent heat of vaporization
Q4 = 966 g × 2260 J/g
Q4 = 2184360 J

Lastly, we calculate the heat required to raise the temperature of steam from 100°C to 126°C. The specific heat capacity of steam is 2.03 J/g°C.

Heat required to raise the temperature of steam from 100°C to 126°C:
Q5 = mass × specific heat capacity × temperature change
Q5 = 966 g × 2.03 J/g°C × (126°C - 100°C)
Q5 = 58680 J

Now, we can sum up all the heat values to obtain the total heat required:

Total heat = Q1 + Q2 + Q3 + Q4 + Q5
Total heat = 20266.8 J + 322644 J + 403925.2 J + 2184360 J + 58680 J
Total heat = 3053875 J

Therefore, approximately 3,053,875 Joules (J) of heat is needed to convert 966 grams (g) of ice at -10°C to steam at 126°C.