Calculate the amount of heat needed to convert 45g of ice at –22.0°C to water at 22.0°C.

q1 = heat needed to raise T of ice from -22 to zero C.

q1 = mass ice x specific heat ice x (Tfinal-Tinitial)

q2 = heat needed to change solid ice at zero C to liquid H2O at zero C.
q2 = mass ice x heat fusion

q3 = heat needed to raise T of liquid water from zero C to 22.0 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial)

Total heat = sum q1+q2+q3.

To calculate the amount of heat needed to convert ice to water, we need to consider two parts: the heat required to raise the temperature of the ice from -22.0°C to 0°C (the melting point of ice), and the heat required to melt the ice to water at 0°C. After that, we calculate the heat required to raise the temperature of the resulting water from 0°C to 22.0°C.

Let's break it down step by step:

Step 1: Calculate the heat required to raise the temperature of the ice from -22.0°C to 0°C.
To find this, we use the specific heat capacity formula:

Q = m * ΔT * C

where:
Q = heat energy (in Joules)
m = mass of the substance (in grams)
ΔT = change in temperature (in Celsius)
C = specific heat capacity (in J/g°C)

The specific heat capacity of ice is approximately 2.09 J/g°C.

Q1 = 45g * (0°C - (-22.0°C)) * 2.09 J/g°C
= 45g * 22.0°C * 2.09 J/g°C
= 2,079.9 J

Step 2: Calculate the heat required to melt the ice to water at 0°C.
The heat required to melt ice is given by the formula:

Q2 = m * L
where:
Q2 = heat energy (in Joules)
m = mass of the substance (in grams)
L = heat of fusion for ice (fusion enthalpy) (in J/g)

The heat of fusion for ice is approximately 333.5 J/g.

Q2 = 45g * 333.5 J/g
= 15,007.5 J

Step 3: Calculate the heat required to raise the temperature of water from 0°C to 22.0°C.
To find this, we use the specific heat capacity formula again.

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

Q3 = 45g * (22.0°C - 0°C) * 4.18 J/g°C
= 45g * 22.0°C * 4.18 J/g°C
= 4,204.2 J

Finally, we add up the three parts to get the total heat energy required:

Total heat energy = Q1 + Q2 + Q3
= 2,079.9 J + 15,007.5 J + 4,204.2 J
= 21,291.6 J

Therefore, it takes approximately 21,291.6 Joules of heat energy to convert 45g of ice at -22.0°C to water at 22.0°C.

To calculate the amount of heat needed to convert ice to water, we need to consider two steps:

1. Heating the ice from -22.0°C to 0°C
2. Melting the ice at 0°C
3. Heating the water from 0°C to 22.0°C

Let's start with step 1:

Step 1: Heating the ice from -22.0°C to 0°C
The specific heat capacity of ice is 2.09 J/g°C.
The temperature change is: 0°C - (-22.0°C) = 22.0°C

Q1 = m * c * ΔT
Q1 = 45g * 2.09 J/g°C * 22.0°C
Q1 = 2079.9 J

So, the heat required to heat the ice from -22.0°C to 0°C is 2079.9 J.

Step 2: Melting the ice at 0°C
The heat of fusion for ice is 334 J/g.

Q2 = m * ΔHf
Q2 = 45g * 334 J/g
Q2 = 15030 J

So, the heat required to melt the ice is 15030 J.

Step 3: Heating the water from 0°C to 22.0°C
The specific heat capacity of water is 4.18 J/g°C.
The temperature change is: 22.0°C - 0°C = 22.0°C

Q3 = m * c * ΔT
Q3 = 45g * 4.18 J/g°C * 22.0°C
Q3 = 4149.9 J

So, the heat required to heat the water from 0°C to 22.0°C is 4149.9 J.

Finally, to calculate the total heat required, we add together the heat from each step:

Total heat = Q1 + Q2 + Q3
Total heat = 2079.9 J + 15030 J + 4149.9 J
Total heat = 21259.8 J

Therefore, the amount of heat needed to convert 45g of ice at -22.0°C to water at 22.0°C is 21259.8 J.