A 149.6-g sample of a metal at 74.4°C is added to 149.6 g H₂O at 15.4°C. The temperature of the water rises to 18.6°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat capacity of water is 4.18 J/°C.g. Specific heat capacity= 3/°C-g Submit Answer Try Another Version item attempt remaining

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### Specific Heat Capacity Calculation

**Problem Statement:**
A 149.6-g sample of a metal at 74.4°C is added to 149.6 g of H₂O at 15.4°C. The temperature of the water rises to 18.6°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat capacity of water is 4.18 J/°C·g.

**Solution Steps:**
1. **Identify the given values:**
   - Mass of the metal (mₘ): 149.6 g
   - Initial temperature of the metal (Tₘ_initial): 74.4°C
   - Final temperature of the metal (Tₘ_final): 18.6°C
   - Mass of water (mₓ): 149.6 g
   - Initial temperature of water (Tₓ_initial): 15.4°C
   - Final temperature of water (Tₓ_final): 18.6°C
   - Specific heat capacity of water (Cₓ): 4.18 J/°C·g

2. **Determine the change in temperature:**
   - Change in temperature of the metal (ΔTₘ): Tₘ_final - Tₘ_initial = 18.6°C - 74.4°C
   - Change in temperature of the water (ΔTₓ): Tₓ_final - Tₓ_initial = 18.6°C - 15.4°C

3. **Calculate the heat gained by the water:**
   - qₓ = mₓ * Cₓ * ΔTₓ
   - Here, mₓ = 149.6 g, Cₓ = 4.18 J/°C·g, ΔTₓ = 18.6°C - 15.4°C = 3.2°C
   - qₓ = 149.6 g * 4.18 J/°C·g * 3.2°C

4. **Calculate the heat lost by the metal:**
   - The heat lost by the metal (qₘ) is equal to the heat gained by the water (qₓ) based on the assumption.
   - qₘ = qₓ = 149.6 g *
Transcribed Image Text:### Specific Heat Capacity Calculation **Problem Statement:** A 149.6-g sample of a metal at 74.4°C is added to 149.6 g of H₂O at 15.4°C. The temperature of the water rises to 18.6°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat capacity of water is 4.18 J/°C·g. **Solution Steps:** 1. **Identify the given values:** - Mass of the metal (mₘ): 149.6 g - Initial temperature of the metal (Tₘ_initial): 74.4°C - Final temperature of the metal (Tₘ_final): 18.6°C - Mass of water (mₓ): 149.6 g - Initial temperature of water (Tₓ_initial): 15.4°C - Final temperature of water (Tₓ_final): 18.6°C - Specific heat capacity of water (Cₓ): 4.18 J/°C·g 2. **Determine the change in temperature:** - Change in temperature of the metal (ΔTₘ): Tₘ_final - Tₘ_initial = 18.6°C - 74.4°C - Change in temperature of the water (ΔTₓ): Tₓ_final - Tₓ_initial = 18.6°C - 15.4°C 3. **Calculate the heat gained by the water:** - qₓ = mₓ * Cₓ * ΔTₓ - Here, mₓ = 149.6 g, Cₓ = 4.18 J/°C·g, ΔTₓ = 18.6°C - 15.4°C = 3.2°C - qₓ = 149.6 g * 4.18 J/°C·g * 3.2°C 4. **Calculate the heat lost by the metal:** - The heat lost by the metal (qₘ) is equal to the heat gained by the water (qₓ) based on the assumption. - qₘ = qₓ = 149.6 g *
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