Question:
A 100 g metal at 80°C is placed in 100 g water at 20°C. Final temperature is 40°C. Find specific heat of metal.
A 100 g metal at 80°C is placed in 100 g water at 20°C. Final temperature is 40°C. Find specific heat of metal.
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Since the masses of the two substances are equal ($100$ g each), you can simplify the calculation by canceling them out from both sides immediately. The ratio of the specific heats will be inversely proportional to the ratio of their temperature changes.
Updated On: Jun 3, 2026
- 420 J/kgK
- 840 J/kgK
- 1680 J/kgK
- 2100 J/kgK
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The Correct Option is D
Solution and Explanation
Concept:
According to the Principle of Calorimetry, when two bodies at different temperatures are placed in contact, the heat lost by the hot body is equal to the heat gained by the cold body, provided no heat is lost to the surroundings.
Step 1: Identify the parameters
• For Metal (Hot body):
• Mass ($m_m$) = 100 g = 0.1 kg
• Initial Temperature ($T_m$) = 80°C
• Final Temperature ($T_f$) = 40°C
• Specific heat ($s_m$) = ?
• For Water (Cold body):
• Mass ($m_w$) = 100 g = 0.1 kg
• Initial Temperature ($T_w$) = 20°C
• Final Temperature ($T_f$) = 40°C
• Specific heat of water ($s_w$) = 4200 J/kgK (Standard value)
Step 2: Apply the heat balance equation.
\[ \text{Heat Lost by Metal} = \text{Heat Gained by Water} \] \[ m_m \cdot s_m \cdot (T_m - T_f) = m_w \cdot s_w \cdot (T_f - T_w) \] Substituting the values: \[ 0.1 \cdot s_m \cdot (80 - 40) = 0.1 \cdot 4200 \cdot (40 - 20) \] \[ s_m \cdot 40 = 4200 \cdot 20 \]
Step 3: Solve for $s_m$.
\[ s_m = \frac{4200 \cdot 20}{40} \] \[ s_m = \frac{4200}{2} \] \[ s_m = 2100 \text{ J/kgK} \]
Step 1: Identify the parameters
• For Metal (Hot body):
• Mass ($m_m$) = 100 g = 0.1 kg
• Initial Temperature ($T_m$) = 80°C
• Final Temperature ($T_f$) = 40°C
• Specific heat ($s_m$) = ?
• For Water (Cold body):
• Mass ($m_w$) = 100 g = 0.1 kg
• Initial Temperature ($T_w$) = 20°C
• Final Temperature ($T_f$) = 40°C
• Specific heat of water ($s_w$) = 4200 J/kgK (Standard value)
Step 2: Apply the heat balance equation.
\[ \text{Heat Lost by Metal} = \text{Heat Gained by Water} \] \[ m_m \cdot s_m \cdot (T_m - T_f) = m_w \cdot s_w \cdot (T_f - T_w) \] Substituting the values: \[ 0.1 \cdot s_m \cdot (80 - 40) = 0.1 \cdot 4200 \cdot (40 - 20) \] \[ s_m \cdot 40 = 4200 \cdot 20 \]
Step 3: Solve for $s_m$.
\[ s_m = \frac{4200 \cdot 20}{40} \] \[ s_m = \frac{4200}{2} \] \[ s_m = 2100 \text{ J/kgK} \]
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