Specific Heat Capacity

What you’ll learn

Calculate energy for a temperature change without a phase change.

គណនាថាមពលកម្ដៅសម្រាប់ការប្ដូរសីតុណ្ហភាពដែលគ្មានប្ដូរសភាព។

Meaning and use

Specific heat capacity relates energy input to mass and temperature change. Use the final minus initial temperature, not the final temperature alone. The simple formula assumes a suitable approximately constant c and no phase change. A Celsius temperature difference has the same numerical size as a kelvin difference.

កម្ដៅម៉ាសភ្ជាប់ថាមពល ម៉ាស និងបម្រែបម្រួលសីតុណ្ហភាព។ ប្រើសីតុណ្ហភាពចុងដកដើម។ រូបមន្តនេះប្រើពេលគ្មានប្ដូរសភាព និង c ប្រហែលថេរ។

Formulas

Energy for temperature change

\[Q=mc\Delta T\]

Temperature difference

\[\Delta T=T_f-T_i\]

Symbols and units

Symbol	Meaning	Unit or restriction
Q	Energy transferred as heat to the stated material	J
m	Mass	kg
c	Specific heat capacity	J/(kg·K)
ΔT	Temperature change	K or a temperature difference in °C

Conditions and limits

• No melting, boiling, or other phase change occurs during the calculation.
• Use the supplied c; examples take water c = 4200 J/(kg·K).
• បើមានការបាត់កម្ដៅទៅបរិស្ថាន ថាមពលឧបករណ៍បញ្ចូលអាចលើស Q នៃវត្ថុ។

Worked examples

Heating water

Heat 0.5 kg water from 20°C to 30°C, c = 4200 J/(kg·K).

1. Temperature rise \(\Delta T=30-20=10\) K.
2. Calculate \(Q=0.5\times4200\times10=21000\) J.

21000 J

ត្រូវប្រើការកើន 10 ដឺក្រេ មិនមែន 30។

Finding a temperature rise

A 2 kg block has c = 500 J/(kg·K) and receives 10000 J, with no other transfer.

1. Rearrange \(\Delta T=Q/(mc)\).
2. Calculate \(10000/(2\times500)=10\) K.

10 K temperature rise

ចែកថាមពលនឹងផលគុណម៉ាសនិងកម្ដៅម៉ាស។

Common mistakes

Instead of	Use	Why
Q = mcTf in every heating problem.	Q = mc(Tf − Ti).	Use the change from the initial temperature.
Use Q = mcΔT to melt material at constant temperature.	Use latent heat for the phase change.	The energy-temperature relationship differs during melting.

Practice

Solve these before revealing the worked solutions. Include units where relevant.

សាកល្បងដោះស្រាយមុនបើកមើលចម្លើយ។ សូមបង្ហាញជំហាន និងឯកតាដែលពាក់ព័ន្ធ។

1. Find Q for m = 1 kg, c = 1000 J/(kg·K), ΔT = 5 K.
2. Water warms from 15°C to 25°C. Find ΔT.
3. If mass doubles for the same c and ΔT, how does Q change?

Answers and steps

Show solutions / បង្ហាញដំណោះស្រាយ

1. 5000 J

   1 × 1000 × 5 = 5000.

2. 10 K, equivalently a 10°C difference

   Subtract the initial value from the final value.

3. It doubles.

   Q is proportional to mass in this model.

Quick recap

• Use temperature change, not final temperature.
• Check that no phase change occurs.
• Use consistent mass and heat-capacity units.