Question

A metal ball of mass 100 g at \( 20^\circ C \) is dropped in 200 ml of water at \( 80^\circ C \). If the resultant temperature is \( 70^\circ C \), then the ratio of specific heat of the metal to that of water is:

• A. \( \frac{5}{2} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{2}{1} \)

Hint:

In calorimetry problems, use the principle "Heat lost = Heat gained" and ensure correct unit conversions for mass and temperature changes.

Answer 1

The Correct Option is C

Step 1: Applying the Principle of Heat Exchange According to the principle of calorimetry: \[ \text{Heat lost by hot body} = \text{Heat gained by cold body}. \] For the given system: - The metal ball gains heat. - The water loses heat.
Step 2: Writing the Heat Exchange Equations The heat gained by the metal is: \[ Q_{\text{metal}} = m_{\text{metal}} c_{\text{metal}} \Delta T_{\text{metal}}. \] The heat lost by water is: \[ Q_{\text{water}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}}. \] Since heat lost = heat gained: \[ m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}} = m_{\text{metal}} c_{\text{metal}} \Delta T_{\text{metal}}. \]
Step 3: Substituting the Given Values - \( m_{\text{metal}} = 100 \) g = 0.1 kg. - \( m_{\text{water}} = 200 \) ml = 0.2 kg (since density of water is 1000 kg/m³). - Specific heat of water, \( c_{\text{water}} = 1 \) (in relative units). - Temperature changes: \[ \Delta T_{\text{metal}} = 70^\circ C - 20^\circ C = 50^\circ C. \] \[ \Delta T_{\text{water}} = 80^\circ C - 70^\circ C = 10^\circ C. \]
Step 4: Solving for \( c_{\text{metal}} \) \[ (0.2)(1)(10) = (0.1)(c_{\text{metal}})(50). \] \[ 2 = 5 c_{\text{metal}}. \] \[ c_{\text{metal}} = \frac{2}{5}. \]
Step 5: Finding the Ratio of Specific Heats The ratio of the specific heat of the metal to that of water is: \[ \frac{c_{\text{metal}}}{c_{\text{water}}} = \frac{2}{5}. \] Thus, the correct answer is: \[ \boxed{\frac{2}{5}}. \]