A 50 g of a metal absorbed 2000 J of heat energy which raised the temperature by 10°C Calculate the specific heat capacity of the metal?

Answer

Specific Heat Capacity Calculation | Thermodynamics Example

🔥 Specific Heat Capacity Calculation

🧾 Step 1: Introduction

Specific heat capacity (c) is the amount of heat required to raise the temperature of 1 kg of a substance by 1°C. It is commonly used in material science and thermodynamics to understand heat transfer.

📌 Step 2: Heat Energy Formula

Q = mcΔT
  • Q = Heat energy (Joules)
  • m = Mass (kg)
  • c = Specific heat capacity (J/kg·°C)
  • ΔT = Temperature change (°C)

📊 Step 3: Given Data

Parameter Value
Mass of metal 50 g = 0.050 kg
Heat absorbed 2000 J
Temperature rise 10°C

🧠 Step 4: Rearranging the Formula

To solve for c, rearrange the formula:

c = Q / (m × ΔT)

🔢 Step 5: Substitute Values

c = 2000 / (0.050 × 10) = 2000 / 0.5 = 4000 J/kg·°C
Final Answer: The specific heat capacity of the metal is 4000 J/kg·°C.

📘 Conclusion

This result means that 4000 joules of energy is required to raise the temperature of 1 kg of the given metal by 1°C. This information is useful in designing thermal systems like:

  • Cookware and kitchen utensils 🍳
  • Heat sinks and exchangers 🌡️
  • Engines and motors 🔧
  • Thermal insulation materials 🧊

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