“How To Solve It” by George Polya, a classic book published in 1945, suggests steps in solving a problem – mathematical or otherwise. The book includes advice for teaching students of mathematics and a mini-encyclopaedia of heuristic terms. Teachers and parents would find it easy to explain problem solving discipline to children. Students will benefit by using the simple framework and tips from Polya. In schools, mathematics question papers contain only problems. In science, a sure way to score better is by attempting all the numerical problems in question papers.

I strongly suggest that parents, teachers and students study “How To Solve It.”

While teaching a few lessons to high school students, I used a modified approach and found it simple and effective. I have tried this approach to teach solving problems in a chapter on heat in Physics text book and a chapter on Equity Shares in Mathematics text book. Let us trace the steps in solving a problem on heat.

Preparatory step:

- Identify the chapter of the problem.
- List all the formulae in the chapter. A problem from the chapter would certainly use the formulae partly or fully.

E.g., in the chapter on Heat, we have the following formulas:

Calorie = 4.186 Joules

ΔQ = c m ΔT (change in heat energy = specific heat of the material X mass X change in temperature)

C’ = ΔQ / ΔT (heat capacity of material = change in heat energy / change in temperature)

c = C’ / m = ΔQ / m ΔT

1 Watt electrical power for 1 second = 1 Joule

L = ΔQ / m (latent heat = change in heat energy / mass)

3. Make a table with one column per variable in the set of formulae. Include the units of measure for each variable, in SI units. In the table, t1 is the starting temperature and t2 is the ending temperature of the process in a step.

m Kg | c : J/(kg*K) | ΔQ J | t1 ^{O}C |
t2 ^{O}C |
ΔT | L J/Kg |

Understand the Problem

The following diagram summarises the theory the student has studied in the chapter on Heat.

Example Problem: Calculate the total amount of heat energy required to convert 100 g of ice at -10^{O}C completely into water vapour at 150^{O}C.

4. Values in the problem would map to one of the variables in the formulae. Make one row per sample item in the problem and enter the values under appropriate variables.

In the problem above, we have 5 samples representing each phase of the conversion of sub-zero ice and super-heated steam: heating ice, melting ice, heating water, vaporising water, and heating steam. Thus we would have 5 rows. Add a row for Total values as the problem requires us to find the total energy.

m Kg | c : J/(kg*K) | ΔQ J | t1 ^{O}C |
t2 ^{O}C |
ΔT | L | |

heating ice | 0.1 | 2,100 | -10 | 0 | |||

melting ice | 0.1 | 0 | 0 | 3,36,000 | |||

heating water | 0.1 | 4,200 | 0 | 100 | |||

vaporising water | 0.1 | 0 | 0 | 22,60,000 | |||

heating steam | 0.1 | 2,000 | 100 | 150 | |||

Total | |

Values of specific heats and latent heats may be given in the problem or the student may be expected to know them.

** **Important: If the problem has given values in different units, convert to SI units. This applies to values of specific heat and latent heat in the current problem. A common cause of error is using inconsistent units for variables

**5. **Picking up values of different variables from the table, calculate and fill values in empty cells.

– I have made a copy of the table to demonstrate the steps. In practice, we would fill the values in the table of the previous step.

– Here, I show the derived values in blue. While working on paper, we may encircle the interim results from solution steps.

– From t1 and t2, calculate ΔT for each row.

– Energy required to bring the temperature of heating ice from 10 to 0^{ O}C

ΔQ = c m ΔT = 2,100 * 0.1 * 10 = 2,100 J

– Energy to convert melting ice to water at 0^{O}C

ΔQ = m L = 0.1 * 3,36,000 = 33,600 J

– Energy to heat water from 0^{O}C to 100^{O}C

ΔQ = c m ΔT = 4,200 * 0.1 * 100 = 42,000 J

– Energy to vaporise the water at 100^{O}C

ΔQ = m L = 0.1 * 22,60,000 = 2,26,000 J

– Energy to heat steam from 100^{O}C to 150^{O}C

ΔQ = c m ΔT = 2,000 * 0.1 * 50 = 10,000 J

m Kg | c : J/(kg*K) | ΔQ J | t1 ^{O}C |
t2 ^{O}C |
ΔT | L | |

heating ice | 0.1 | 2,100 | 2,100 | -10 | 0 | 10 | |

melting ice | 0.1 | 33,600 | 0 | 0 | 0 | 3,36,000 | |

heating water | 0.1 | 4,200 | 42,000 | 0 | 100 | 100 | |

vaporising water | 0.1 | 2,26,000 | 0 | 0 | 0 | 22,60,000 | |

heating steam | 0.1 | 2,000 | 10,000 | 100 | 150 | 50 | |

Total | 3,13,700 |

**6. **From the interim results, calculate the final answer and insert in the table. On paper, we could underline the final answer, which is shown in red here.

Total energy spent = 3,13,700 J

7. Review the answer. Before re-calculating, compare the relative energy spent in various phases, recorded in ΔQ column, and check if they conform to the relative lengths of X axis in the picture for different phases.

With a different approach, we have used tables to teach and summarise lessons from other subjects like History and languages. We can discuss about them in another blog.

In the mean time, if you would try this approach as parents, teachers, students or out of curiosity, and give feedback at bhat.pg@gmail.com , I shall be grateful. The feedback will help refine the approach.

The approach i recollect having used in school didn’t use tables but was similar in approach. Example, delQ = m*c*delT

Given information:

delQ?

m = 0.1 KG

c = 2400 J/(Kg*K)

delT = 10K

Substituting into, formula, we get delQ = 0.1*2400*10 = 2400J

I wonder if you agree that the tabular approach is similar. Or does it do anything different?

Also, I agree that the tabulation approach might help structure the information for students.The key with new learning methodologies is that it is worth testing with a set of students. Example, 2 sections of the same class in the same school can use two different approaches (9A Vs 9B). One gets taught the regular method and another the tabular approach. One can get concrete data on the impact. I imagine it’s hard to rope in teachers willing to experiment. In its absence, some anecdotal evidence would be useful.

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Thanks, Kartik.

There is no difference in the steps in solving the problem. I hope that the tables would help in visualising the relations by putting all the data in one place. It could prompt to convert all the values to SI units.

Informal feedback from the students and teachers was encouraging. That could also be out of their personal regard for me. I have to try this in a school that does not know me.

Now it is very close to the final examination for the students. I would like to try this approach on one section of a class and then compare with the other that does not use the approach.

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