Heat Capacity

Question

A 50.0 g sample of water is heated from 25.0°C to 45.0°C. (Specific heat of water = 4.18 J g⁻¹ K⁻¹). Heat absorbed (q) is:

RankGuru · Accepted Answer

1. **Formula**: The heat absorbed or released during a temperature change is given by $q = m \cdot c \cdot \Delta T$.
2. **Variables**:
   - Mass ($m$) = $50.0$ g.
   - Specific heat capacity ($c$) = $4.18 	ext{ J g}^{-1} 	ext{ K}^{-1}$.
   - Temperature change ($\Delta T$) = $T_{	ext{final}} - T_{	ext{initial}} = 45.0 - 25.0 = 20.0^\circ C$ (or $20.0$ K).
3. **Calculation**:
   - $q = 50.0 	imes 4.18 	imes 20.0$
   - $q = 1000 	imes 4.18 / 1$
   - $q = 4180$ J.
4. **Result**: The energy absorbed by the water is **$4180$ J**.

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