A silver cube with an edge length of 2.24 cm and a gold Substance Specific heat (J/g•°C) Density (g/cm³) cube with an edge length of 2.63 cm are both heated to gold 0.1256 19.3 84.8 °C and placed in 115.0 mL of water at 20.5 °C . silver 0.2386 10.5 What is the final temperature of the water when thermal water 4.184 1.00 equilibrium is reached? Tinal = °C
A silver cube with an edge length of 2.24 cm and a gold Substance Specific heat (J/g•°C) Density (g/cm³) cube with an edge length of 2.63 cm are both heated to gold 0.1256 19.3 84.8 °C and placed in 115.0 mL of water at 20.5 °C . silver 0.2386 10.5 What is the final temperature of the water when thermal water 4.184 1.00 equilibrium is reached? Tinal = °C
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Problem Statement
A silver cube with an edge length of 2.24 cm and a gold cube with an edge length of 2.63 cm are both heated to 84.8 °C and placed in 115.0 mL of water at 20.5 °C. What is the final temperature of the water when thermal equilibrium is reached?
### Data Table
| Substance | Specific Heat (J/g°C) | Density (g/cm³) |
|-----------|-----------------------|-----------------|
| Gold | 0.1256 | 19.3 |
| Silver | 0.2386 | 10.5 |
| Water | 4.184 | 1.00 |
### Calculation Setup
The goal is to find the final temperature (\( T_{\text{final}} \)) of the system when the cubes and water reach thermal equilibrium.
#### Steps:
1. Calculate the mass of each cube using their volumes and densities.
2. Apply the principle of conservation of energy, where the heat lost by the metals equals the heat gained by the water.
3. Use the specific heat capacity values given in the table to calculate the energy exchange.
### Formula
\[ q_{\text{lost by metals}} = q_{\text{gained by water}} \]
\[ m_{\text{gold}} \cdot c_{\text{gold}} \cdot (T_{\text{initial gold}} - T_{\text{final}}) + m_{\text{silver}} \cdot c_{\text{silver}} \cdot (T_{\text{initial silver}} - T_{\text{final}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial water}}) \]
#### Note:
- \( m \) represents mass.
- \( c \) represents specific heat.
Fill in the \( T_{\text{final}} \) box once calculated.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9985dfa2-26db-459e-98b3-af2531f0e329%2Fe4906fa2-243b-4115-9de7-bd4d46bd1817%2F6aafxmm_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A silver cube with an edge length of 2.24 cm and a gold cube with an edge length of 2.63 cm are both heated to 84.8 °C and placed in 115.0 mL of water at 20.5 °C. What is the final temperature of the water when thermal equilibrium is reached?
### Data Table
| Substance | Specific Heat (J/g°C) | Density (g/cm³) |
|-----------|-----------------------|-----------------|
| Gold | 0.1256 | 19.3 |
| Silver | 0.2386 | 10.5 |
| Water | 4.184 | 1.00 |
### Calculation Setup
The goal is to find the final temperature (\( T_{\text{final}} \)) of the system when the cubes and water reach thermal equilibrium.
#### Steps:
1. Calculate the mass of each cube using their volumes and densities.
2. Apply the principle of conservation of energy, where the heat lost by the metals equals the heat gained by the water.
3. Use the specific heat capacity values given in the table to calculate the energy exchange.
### Formula
\[ q_{\text{lost by metals}} = q_{\text{gained by water}} \]
\[ m_{\text{gold}} \cdot c_{\text{gold}} \cdot (T_{\text{initial gold}} - T_{\text{final}}) + m_{\text{silver}} \cdot c_{\text{silver}} \cdot (T_{\text{initial silver}} - T_{\text{final}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial water}}) \]
#### Note:
- \( m \) represents mass.
- \( c \) represents specific heat.
Fill in the \( T_{\text{final}} \) box once calculated.
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