Question

Two bodies of same mass having temperature $T_1$ and $T_2$ place kept contact. Specific heat capacity of bodies are s and 1.2s respectively. find equilibrium temperature.

Hint:

The equilibrium temperature is a weighted average: $T_{eq} = \frac{m_1s_1T_1 + m_2s_2T_2}{m_1s_1 + m_2s_2}$. If masses are equal, it simplifies to $\frac{s_1T_1 + s_2T_2}{s_1 + s_2}$.

Answer 1

Step 1: Understanding the Concept:
When two bodies at different temperatures are placed in thermal contact in an isolated system, heat flows from the hotter body to the colder body until they reach a common equilibrium temperature. The principle of calorimetry states that heat lost equals heat gained.

Step 2: Key Formula or Approach:
Heat transferred $Q = m \cdot s \cdot \Delta T$.
Set $\text{Heat Lost} = \text{Heat Gained}$.

Step 3: Detailed Explanation:
Let the two bodies be A and B.
Mass of A = $m_A = m$, Specific heat $s_A = s$, Initial temp = $T_1$
Mass of B = $m_B = m$, Specific heat $s_B = 1.2s$, Initial temp = $T_2$
Let the final equilibrium temperature be $T$.
Assume $T_1>T_2$. Body A loses heat, Body B gains heat.
Heat lost by A: $Q_{\text{lost}} = m_A \cdot s_A \cdot (T_1 - T) = m \cdot s \cdot (T_1 - T)$
Heat gained by B: $Q_{\text{gained}} = m_B \cdot s_B \cdot (T - T_2) = m \cdot (1.2s) \cdot (T - T_2)$
According to calorimetry principle:
\[ Q_{\text{lost}} = Q_{\text{gained}} \]
\[ m \cdot s \cdot (T_1 - T) = m \cdot 1.2s \cdot (T - T_2) \]
Cancel common terms $m$ and $s$:
\[ T_1 - T = 1.2(T - T_2) \]
\[ T_1 - T = 1.2T - 1.2T_2 \]
Rearrange to solve for $T$:
\[ T_1 + 1.2T_2 = 1.2T + T \]
\[ T_1 + 1.2T_2 = 2.2T \]
\[ T = \frac{T_1 + 1.2T_2}{2.2} \]

Step 4: Final Answer:
The equilibrium temperature is $\frac{T_1 + 1.2T_2}{2.2}$.