 ## July 19, 2012

### Heat Energy Numerical Problems for EM&T Students

Following are some of the solved Heat / Energy related numerical problems for EM&T students.

Constant / Values we should know:
• Specific heat of Ice, C(ice) = 2060 J/KgK
• Specific heat of Water, C(water) = 4186 J/KgK
• Specific heat of Steam, C(steam) = 1870 J/KgK
• Latent heat of Ice Fusion (Melting), L(fusion) = 334 KJ/Kg
• Latent heat of Ice Vaporization (Steam), L(steam) = 2260 KJ/Kg

Numerical Problem # 1: How many Joule and BTU of heat energy is required to change 4.5 Kg of of ice at -15°C to steam at 121°C ?

SolutionIn this problem we have observed that the water changed its phases or state from ice (at -10°C) to steam (at 121°C). So this thermodynamic process involved specific heat and latent heat at different stages.

Now we have to find the values of Heat Energy at all stages (as shown in diagram below click on the diagram for larger view)

• Q = mCΔT (specific heat)
• Q = mL (latent heat)
As we know:
• Heat Energy = Q
• Similarly total Heat Energy = Q(total) = Q1 + Q2 + Q3 + Q4 + Q5
So give Data is:
• Mass, m = 4.5 Kg
• Temp, T from -15 to 121
Heat Energy Calculation:
1. Q1 = mC(ice)ΔT => 4.5 x 2060 x 15 = 139050 Joules (* ΔT = 15 which is difference between -15°C & 0°C)
2. Q2 = mL(fusion) => 4.5 x 334 K = 150300 Joules (* L(fusion) = 334 x 1000 J/Kg and k = 1000)
3. Q3 = mC(water)ΔT => 4.5 x 4186 x 100 = 1883700 Joules (* ΔT = 100 which is difference between 0°C & 100°C)
4. Q4 = mL(steam)ΔT => 4.5 x 2260 K = 10170000 Joules (* L(steam) = 2260 x 1000 J/Kg and k = 1000)
5. Q5 = mC(steam)ΔT => 4.5 x 1870 x 21 = 176715 Joules (* ΔT = 21 which is difference between 100°C & 121°C)
Adding Q1, Q2, Q3, Q4 & Q5:

139050 + 150300 + 1883700 + 10170000 + 176715 = 13872465 Joules
or
13.8 x 10^6 Joules

To convert Joules in BTU we divide the answer by 1055 because 1 BTU is equal to 1055 Joules

• Q = 13.8 Mega Joules
• Q = 13149.2 BTU Numerical Problem # 2: A rapidly spinning pedal wheel raises the temperature of 20 ml of water from 21°C to 25°C. Calculate the work done in the thermodynamic process. Calculate the heat transferred.
Use C(water) = 4200 J/Kg.

Solution:

Part A) Finding work Done:

As Work Done is equal to Energy so we have to find heat energy

To find heat energy we use Q = mCΔT formula

Given Data is:
• Quantity of water = 20 ml so Mass, m = 0.2 Kg
• (* as 1 ml = 1 g so 200 ml = 200 g and 200 grams are equal to 0.2 Kg)
• ΔT = 4°C (* difference between 25 and 21 is 4)
• C(water) = 4200 J/Kg
Putting values in formula: Q = mC(water)ΔT => 0.2 x 4200 x 4 = 3360 Joules

So Work Done in this thermodynamics process is 3360 Joules

Part B) Calculation of transfer of heat:

Answer: No Heat is transferred in this process. It is simply conversation of energy from one form to another form which is mechanical energy into heat energy. Numerical Problem # 3: An ice cube having mass of 50 grams and initial temperature of -10°C is dropped in 400 gram of water at 40°C (temp. of water). What is the final state, mass and temperature of the mixture if the effects of the container can be neglected ?

Solution:

Given Data:
• Mass of Ice Cube, m(i) = 50 g = 0.05 Kg
• Mass of Water, m(w) = 400 g = 0.4 Kg
• Temperature of Ice Cube, T(ice) = -10°C
• Temperature of Water, T(water) = 40°C
Using law of conversation of energy: "Energy cannot be created nor be destroyed, it only changes its state".

So heat loss by the water is equal to heat gain by the ice (Utilizing law of conversation of energy)

Similarly:
Heat Loss (energy loss) = Heat Gain (energy gain)
So:
Q(a) = Q(b)

(* where: Q(a) = Heat Energy of Water and Q(b) = Heat Energy of Ice)

We assume the final temperature of  water (after ice cube is dissolved in it) T(f) is greater than 0°C and less than 40°C.

Part a.) Finding the heat energy Q(a) of water by utilizing (Q = mCΔT)

Q(a) = mC(water)ΔT

Q(a) = (0.4)(4186)(T(i) - T(f)) ---------------------- (* T(i) = Initial temp. of water which is 40°C)

Q(a) = (1674.4)(40 - T(f)) ---------------------- (* T(f) = The final temp. we have to find)

Q(a) = ( 66976 - 1674.4T(f) ) ----------------------  (equation 1)

Part b.) Finding the heat energy of Q(b) of Ice by utilizing Q = mCΔTQ = mL

Q(b) = Q(b)1 + Q(b)2 = Q(b)3 -------------------- (* In this problem we have observed that the Ice Cube has changed its state from ice (at -10°C) to Water (at T(f)°C). So this thermodynamic process involved specific heat and latent heat at different stages.)

Q(b)1 = mCΔT --- Q(b)2 = mL --- Q(b)3 = mCΔT

So:

Q(b)1 = mC(ice)ΔT ------------------------------ (Specific Heat of Ice form -10°C to 0°C)
Q(b)1 = (0.05)(2060)(T(i) - T(f))
Q(b)1 = (0.05)(2060)(10 - 0)
Q(b)1 =1030 Joules

Q(b)2 = mL(fusion) ------------------------------ (Latent Heat of Ice at 0°C)
Q(b)2 = (0.05 Kg)(334 KJ/Kg)
Q(b)2 = (0.05)(334 x 10^3)
Q(b)2 = 16700 Joules

Q(b)3 = mC(water)ΔT -------------------------- (Specific Heat of Cube when it dissolved in water)
Q(b)3 = (0.05)(4186)(T(f) - T(i))
Q(b)3 = (209.3)(T(f) - 0)
Q(b)3 = 209.3T(f) Joules

Now adding Q(b)1 + Q(b)2 + Q(b)3 to find the value of Q(b)

Q(b) = 1030 + 16700 + 209.3T(f)

So:

Q(b) =  17730 + 209.3T(f) ----------------------  (equation 2)

Now comparing equation 1 and equation 2

Q(a) = Q(b)
66976 - 1674.4T(f) = 17730 + 209.3T(f) -
66976 - 17730 = 209.3T(f) + 1674.4T(f)
49246 = 1883.7T(f)
49246 / 1883.7 = T(f)                 -
and
T(f) = 26.14°C 