Topic: Sublimation and Fusion

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0 C, to steam at 112.0 °C? The heat capacity of ice is 2.09 J/g⋅°C and that of steam is 2.01 J/g⋅°C.

Any help on how to start this is really appreciated.

Step 1: Heat to raise T of ice at -14.0 °C to 0.0 °C

q_1 = m⋅C_s,ice⋅ΔT
q_1 = (13.0 g)(2.09 J/g⋅°C)[0.0 °C - (-14.0 °C)]
q_1 = 380.38 J
q_1 = 0.38038 kJ

Step 2: Heat to melt ice at 0.0 °C

q_2 = n⋅ΔH_fus
q_2 = [13.0 g⋅(1 mol/18.016 g)⋅(6.02 kJ/mol)
q_2 = [13.0 g⋅(1 mol/18.016 g)]⋅(6.02 kJ/mol)
q_2 = 4.343916519 kJ

Step 3: Heat to raise T of water from 0.0 °C to 100.0 °C

q_3 = m⋅C_s,liq⋅ΔT
q_3 = (13.0 g)(4.18 J/g⋅°C)(100.0 °C - 0.0 °C)
q_3 = 5434 J
q_3 = 5.434 kJ

Step 4: Heat to vaporize water at 100.0 °C

q_4 = n⋅ΔH_vap
q_4 = [13.0 g⋅(1 mol/18.016 g)⋅(40.7 kJ/mol)
q_4 = [13.0 g⋅(1 mol/18.016 g)]⋅(40.7 kJ/mol)
q_4 = 29.36833925 kJ

Step 5: Heat to raise T of water from 0.0 °C to 100.0 °C

q_5 = m⋅C_s,steam⋅ΔT
q_5 = (13.0 g)(2.01 J/g⋅°C)(112.0 °C - 100.0 °C)
q_5 = 313.56 J
q_5 = 0.31356 kJ

Total Heat Required

q_1 + q_2 + q_3 + q_4 + q_5
(0.38038 kJ) + (4.343916519 kJ) + (5.434 kJ) + (29.3683392 kJ) + (0.31356 kJ)
39.84019572 kJ

I'm fairly certain I used the correct values for the heat capacities and the heat of fusion and heat of vaporization. Is this correct?

Do this in several steps.

heat to raise T ice at -14 C to zero.
heat to melt ice at zero.
heat to raise T from zero to 100 C.
heat to vaporize water at 100.
heat to raise steam from 100 to 112 C.
The total is the sum of each part above.
Post your work if you get stuck.

The procedure looks ok. I didn't look up the Cp values but I know 4.18 is ok for liquid water. I didn't check the math but I note that you need to watch the number of significant figures. I think 3 s.f. are allowable.

To find the heat required to warm the ice to steam, we need to consider three steps:

1. Heating the ice from -14.0°C to 0°C (melting)
2. Melting the ice at 0°C to water at 0°C (fusion)
3. Heating the water from 0°C to 112.0°C (vaporization)

Let's calculate them step by step:

1. Heating the ice:
To find the heat required to warm the ice from -14.0°C to 0°C, we'll use the heat capacity of ice.
Heat = mass * specific heat capacity * change in temperature
= 13.0 g * 2.09 J/g⋅°C * (0°C - (-14.0°C))

First, we need to convert the mass from grams to kilograms:
Mass = 13.0 g = 0.013 kg

Now we calculate the heat:
Heat = 0.013 kg * 2.09 J/g⋅°C * (0°C - (-14.0°C))

2. Melting the ice:
To find the heat required to melt the ice, we'll use the latent heat of fusion for ice.
Heat = mass * latent heat of fusion
= 13.0 g * latent heat of fusion

3. Heating the water to steam:
To find the heat required to heat the water from 0°C to 112.0°C, we'll use the heat capacity of water.
Heat = mass * specific heat capacity * change in temperature
= 13.0 g * specific heat capacity of water * (112.0°C - 0°C)

Finally, add up the heats from all three steps to get the total heat required to warm the ice to steam.