Calorimetry Mixture

Question

Final temperature when 0.2 kg water at 80°C mixed with 0.3 kg water at 20°C (no loss)?

RankGuru · Accepted Answer

1. **Principle of Calorimetry**: Heat lost by the hot body = Heat gained by the cold body.
2. **Equations**: $m_1 c (T_{high} - T_f) = m_2 c (T_f - T_{low})$. Since both are water, specific heat $c$ cancels out.
3. **Substitute**:
   - $0.2 	imes (80 - T_f) = 0.3 	imes (T_f - 20)$
4. **Solve**:
   - $16 - 0.2 T_f = 0.3 T_f - 6$
   - $16 + 6 = 0.3 T_f + 0.2 T_f$
   - $22 = 0.5 T_f \implies T_f = \frac{22}{0.5} = 44^\circ$C.
5. **Result**: The equilibrium temperature is **$44^\circ$C**.

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Calorimetry Mixture

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