3. What quantity of heat (in kJ) would be required to convert a 2.5-pound block of ice at -4.0 C to steam at 110.0 °C?1 lb = 2.2 kg

Question

could someone please help me on this question please.

Answer 1

Hi Kenneth!

This is a long one. Basically:

Step 1: Find q to heat ice from -4.0 C to 0 C.

Step 2: Find q to melt the ice.

Step 3: Find q to heat water from 0 C to 100 C.

Step 4: Find q to vaporize water.

Step 5: Find q to heat steam from 100 C to 110 C.

Step 6: Add the q‘s required for each step (Steps 1-5).

Prep Work: Determine needed constants & convert the weight to metric

Constants you will need to know:

1)     the specific heat capacities of ice (2.09 J/g C), water (4.184 J/g C), and steam (2.03 J/g C)

Note: Whenever you see units including J/g, that’s equivilant to kJ/kg.

2)     the heat of fusion for water (334 J/g)

3)     the heat of vaporization for water (2,260 J/g)

4)     the conversion factor between pounds & kilograms (1.00 lb= 2.20 kg)

Convert weight of the block of ice to kilograms

Multiplying by the conversion factor of 1 pound = 2.2 kilograms:

2.5 pounds * 2.2 kilograms/pound = 5.5 kilograms

Step 1: Calculate the q required to raise the T of the ice from -4.0 C to 0 C.

This can be calculated using q = mc∆T, where:

q = heat required

m = mass of the ice

c = specific heat capacity of ice = 2.09 J/g C

∆T = change in temperature (for this step only

So the heat required to raise the temperature of the 5.5-kilogram block of ice from -4.0 C to 0 C is:

q₁ = (5.5 kilograms)(2.09 kJ/kg C)(4.0 C) = 45.996 kJ*

*Note: underlined numbers represent the final place value that would be included if this were the final answer and we were concerned with significant figures.

Step 2: Calculate q required to melt the ice at 0 C.

This can be calculated using q = m∆H_fus where:

q = heat required

m = mass of the ice

∆H_fus = heat of fusion for water = 334 J/g

So the heat required to melt the 5.5-kilogram block of ice is: q₂ = (5.5 kg)(334 kJ/kg) = 1837 kJ

Step 3: Calculate q required to raise the T of the water from 0 C to 100 C.

This can be calculated using q = mc∆T, where:

q = heat required

m = mass of the water

c = specific heat capacity of water = 4.184 J/g C

∆T = change in temperature (for this step only)

So the heat required to raise the temperature of the 5.5-kilogram water from 0 C to 100 C is:

q₃ = (5.5 kilograms)(4.184 kJ/kg C)(100.0 C) = 2301.2 kJ

Step 4: Calculate q required to convert the water at 100 C to steam at 100 C.

This can be calculated using q = m∆H_vap where:

q = heat required

m = mass of the ice (in grams!)

∆H_vap = heat of vaporization for water = 2260 J/g

So the heat required to vaporize the 5.5-kilogram water is: q₄ = (5.5 kg)(2260 kJ/kg) = 12430 kJ

Step 5: Calculate q required to convert the steam at 100 C to steam at 110 C.

This can be calculated using q = mc∆T, where:

q = heat required

m = mass of the steam

c = specific heat capacity of steam = 2.03 J/g C

∆T = change in temperature (for this step only)

So the heat required to raise the temperature of the 5.5-kilogram steam from 100 C to 110 C is:

q₅ = (5.5 kilograms)(2.03 kJ/kg C)(10.0 C) = 111.65 kJ

Step 6: Add the q‘s required for each step (Steps 1-5).

Adding the heat required for each step, the total heat required to convert the 2.5-pound block of ice at -4.0 C to steam at 110.0 C is:

45.996 kJ + 1837 kJ + 2301.2 kJ + 12430 kJ + 111.65 kJ = 16725.846 kJ

Therefore, the answer to the question should be: 17000 kJ.

Hopefully this helps!

Answer 2

You have to supply the energy to heat up the different phases to transition temperatures (T_FP=0 and T_BP= 100) and to change phases. You need to obtain c_Ice, c_water, and c_steam and latent heat of fusion and vaporization. You will have to pay attention to units (Put kJ into Joles and pounds have to be converted to kg).

mc_ice(0-(-4))+ mL_Fus + mc_Water(100-0) + mL_Vap + mc_Steam(110-100) = Q in Joules.

If you are careful about units. you should be fine.(if c's are in kJ/kg, you can do everything in kJ)

Please consider a tutor. Take care.