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
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
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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 *](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96f2e08c-aa5d-4493-8570-9d5ecf302bf0%2F3fd96fdd-5014-47aa-9db7-0f95cd0618c6%2F5io0t4p_processed.jpeg&w=3840&q=75)
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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