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

To solve the problem of finding the ratio of the specific heat of the metal to that of water, we need to apply the principle of conservation of energy. When the metal ball is dropped into the water, heat lost by water = heat gained by the metal. Let \( m_m \) and \( c_m \) be the mass and specific heat of the metal, and \( m_w \) and \( c_w \) be the mass and specific heat of water, respectively. Given:

• Mass of the metal ball, \( m_m = 100 \text{ g} = 0.1 \text{ kg} \).
• Initial temperature of the metal ball, \( T_{m, \text{initial}} = 20^\circ C \).
• Mass of water, \( m_w = 200 \text{ ml} = 0.2 \text{ kg} \) (assuming density of water is 1 g/ml).
• Initial temperature of the water, \( T_{w, \text{initial}} = 80^\circ C \).
• Final temperature of the system, \( T_{\text{final}} = 70^\circ C \).

Applying the principle of conservation of energy:

\(m_w \cdot c_w \cdot (T_{w, \text{initial}} - T_{\text{final}}) = m_m \cdot c_m \cdot (T_{\text{final}} - T_{m, \text{initial}})\)

Substituting the known values:

\(0.2 \cdot c_w \cdot (80 - 70) = 0.1 \cdot c_m \cdot (70 - 20)\)

Simplifying the equation gives:

\(0.2 \cdot c_w \cdot 10 = 0.1 \cdot c_m \cdot 50\)

\(2c_w = 5c_m\)

Rearranging for the ratio of specific heat:

\(\frac{c_m}{c_w} = \frac{2}{5}\)

Therefore, the ratio of specific heat of the metal to that of water is \(\frac{2}{5}\).

Answer 2

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: \[ \boxed{\frac{2}{5}}. \]