A 50.0 g sample of Chromium metal was heated to 103 degrees Celcius and then dropped into a beaker containing 76.0 g of water at 25.0 degrees Celcius. If the specific heat of Chromium metal is 0.110 cal/g C, what will be the equilibrium temperature of the Chromium-Water mixture? (Cwater = 1 cal/g °C)

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**Problem Statement:**

A 50.0 g sample of Chromium metal was heated to 103 degrees Celsius and then dropped into a beaker containing 76.0 g of water at 25.0 degrees Celsius. If the specific heat of Chromium metal is 0.110 cal/g°C, what will be the equilibrium temperature of the Chromium-Water mixture? (C_water = 1 cal/g°C)

---

In this physics and chemistry problem, you're asked to find the final equilibrium temperature when a hot metal (Chromium) is mixed with cooler water. Below are the relevant details extracted from the problem for clarity:

- Mass of Chromium (m_Cr): 50.0 g
- Initial temperature of Chromium (T_initial_Cr): 103°C
- Specific heat of Chromium (C_Cr): 0.110 cal/g°C

- Mass of water (m_water): 76.0 g
- Initial temperature of water (T_initial_water): 25°C
- Specific heat of water (C_water): 1 cal/g°C

The goal is to determine the final equilibrium temperature (T_final) of the combined Chromium-Water mixture. 

To solve this, you would typically use the formula for heat transfer and assume that the heat lost by the Chromium equals the heat gained by the water:

\[ q_{Cr} = -q_{water} \]

Where:
- \( q = mc\Delta T \)
- \( \Delta T = T_{final} - T_{initial} \)

### Detailed Solution Process:

1. Write the heat loss equation for Chromium:
\[ q_{Cr} = m_{Cr} \cdot C_{Cr} \cdot (T_{final} - T_{initial_{Cr}}) \]

2. Write the heat gain equation for water:
\[ q_{water} = m_{water} \cdot C_{water} \cdot (T_{final} - T_{initial_{water}}) \]

3. Set the heat loss equal to the heat gain and solve for \( T_{final} \):
\[ m_{Cr} \cdot C_{Cr} \cdot (T_{final} - T_{initial_{Cr}}) = - (m_{water} \cdot C_{water} \cdot (T_{final} - T_{initial_{water}})) \]

### Substitute the given values and solve:

\[ (50.0
Transcribed Image Text:**Problem Statement:** A 50.0 g sample of Chromium metal was heated to 103 degrees Celsius and then dropped into a beaker containing 76.0 g of water at 25.0 degrees Celsius. If the specific heat of Chromium metal is 0.110 cal/g°C, what will be the equilibrium temperature of the Chromium-Water mixture? (C_water = 1 cal/g°C) --- In this physics and chemistry problem, you're asked to find the final equilibrium temperature when a hot metal (Chromium) is mixed with cooler water. Below are the relevant details extracted from the problem for clarity: - Mass of Chromium (m_Cr): 50.0 g - Initial temperature of Chromium (T_initial_Cr): 103°C - Specific heat of Chromium (C_Cr): 0.110 cal/g°C - Mass of water (m_water): 76.0 g - Initial temperature of water (T_initial_water): 25°C - Specific heat of water (C_water): 1 cal/g°C The goal is to determine the final equilibrium temperature (T_final) of the combined Chromium-Water mixture. To solve this, you would typically use the formula for heat transfer and assume that the heat lost by the Chromium equals the heat gained by the water: \[ q_{Cr} = -q_{water} \] Where: - \( q = mc\Delta T \) - \( \Delta T = T_{final} - T_{initial} \) ### Detailed Solution Process: 1. Write the heat loss equation for Chromium: \[ q_{Cr} = m_{Cr} \cdot C_{Cr} \cdot (T_{final} - T_{initial_{Cr}}) \] 2. Write the heat gain equation for water: \[ q_{water} = m_{water} \cdot C_{water} \cdot (T_{final} - T_{initial_{water}}) \] 3. Set the heat loss equal to the heat gain and solve for \( T_{final} \): \[ m_{Cr} \cdot C_{Cr} \cdot (T_{final} - T_{initial_{Cr}}) = - (m_{water} \cdot C_{water} \cdot (T_{final} - T_{initial_{water}})) \] ### Substitute the given values and solve: \[ (50.0
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