**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 ?

**Solution**: In 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)

- Heat Energy = Q
- Similarly total Heat Energy =
**Q**(total) = Q1 + Q2 + Q3 + Q4 + Q5

- Mass, m = 4.5 Kg
- Temp, T from -15 to 121

Heat Energy Calculation:

**Q1****=****mC**(ice)**ΔT**=> 4.5 x 2060 x 15 =**139050 Joules**(***ΔT**= 15*which is difference between -15°C & 0°C*)**Q2**=**mL**(fusion) => 4.5 x 334 K =**150300 Joules**(* L(fusion) = 334 x 1000 J/Kg and k = 1000)**Q3**=**mC**(water)**ΔT**=> 4.5 x 4186 x 100 =**1883700 Joules**(***ΔT**= 100*which is difference between 0°C & 100°C*)**Q4**=**mL**(steam)**ΔT**=> 4.5 x 2260 K =**10170000 Joules**(* L(steam) = 2260 x 1000 J/Kg and k = 1000)**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 =

or

**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**

**Answer**:

**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

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

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 = mC****ΔT**)**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 = mCΔT & Q = 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**

**Answer**:

Total mass of mixture is 0.05 Kg + 0.4 Kg =

**0.45 Kg**

The final state of mixture is water.

The final temperature of mixture is

**26.14°C**