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. 420 J/kgK
• B. 840 J/kgK
• C. 1680 J/kgK
• D. 2100 J/kgK

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem follows the Law of Calorimetry, which states that in an isolated system, the heat lost by a hotter body must equal the heat gained by a cooler body until thermal equilibrium is reached.
Step 2: Key Formula or Approach:
Heat change \( Q = mc \Delta T \).
Balance: \( m_{m} c_{m} (T_{m} - T_{f}) = m_{w} c_{w} (T_{f} - T_{w}) \).
Take \(c_{w} = 4200\) J/kgK (standard specific heat of water).
Step 3: Detailed Explanation:
Metal: \(m_{m} = 0.1\) kg, \(T_{initial} = 80^\circ\)C, \(T_{final} = 40^\circ\)C, \(\Delta T = 40^\circ\)C.
Water: \(m_{w} = 0.1\) kg, \(T_{initial} = 20^\circ\)C, \(T_{final} = 40^\circ\)C, \(\Delta T = 20^\circ\)C.
Heat lost = Heat gained:
\[ 0.1 \times c_{m} \times (80 - 40) = 0.1 \times 4200 \times (40 - 20) \]
Canceling 0.1 from both sides:
\[ c_{m} \times 40 = 4200 \times 20 \]
\[ c_{m} = \frac{4200 \times 20}{40} = 2100 \text{ J/kgK} \]
Step 4: Final Answer:
Specific heat of metal is 2100 J/kgK.