Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/gx°C. O 837 J O 83.7 J O 2.10 x 106 J 47.8 J

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### Heat Calculation in Physics

#### Question:
Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/g×°C.

#### Options:
- 837 J
- 83.7 J
- 2.10 x 10^6 J
- 47.8 J

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- Number of questions saved: 3 out of 30 questions

---

This question is part of a physics quiz designed to assess understanding of the concept of specific heat and the calculation of heat transfer. The specific heat formula used to solve this is:

\[ q = mc\Delta T \]

Where:
- \( q \) is the heat energy (in Joules)
- \( m \) is the mass (in grams)
- \( c \) is the specific heat capacity (in J/g×°C)
- \( \Delta T \) is the change in temperature (in °C)

Given data:
- Mass of water, \( m \) = 20.0 g
- Initial temperature, \( T_i \) = 25.1°C
- Final temperature, \( T_f \) = 35.1°C
- Specific heat capacity of water, \( c \) = 4.184 J/g×°C

The temperature change, \( \Delta T = T_f - T_i = 35.1°C - 25.1°C = 10.0°C \).

Now substitute these values into the formula:

\[ 
q = 20.0 \text{ g} \times 4.184 \text{ J/g} \times 10.0 \text{°C} 
\]

\[ 
q = 837 J 
\]

Therefore, the quantity of heat required is **837 Joules**.

This example illustrates the practical application of the specific heat concept and demonstrates how to perform necessary calculations.
Transcribed Image Text:### Heat Calculation in Physics #### Question: Calculate the quantity of heat (in Joules) required to raise the temperature of 20.0 g of water from 25.1°C to 35.1°C. The specific heat of water is 4.184 J/g×°C. #### Options: - 837 J - 83.7 J - 2.10 x 10^6 J - 47.8 J #### Navigation: - **Previous Page**: [Button for navigating to the previous page] - **Next Page**: [Button for navigating to the next page] #### Quiz Control: - **Submit Quiz**: [Button to submit the quiz] - Number of questions saved: 3 out of 30 questions --- This question is part of a physics quiz designed to assess understanding of the concept of specific heat and the calculation of heat transfer. The specific heat formula used to solve this is: \[ q = mc\Delta T \] Where: - \( q \) is the heat energy (in Joules) - \( m \) is the mass (in grams) - \( c \) is the specific heat capacity (in J/g×°C) - \( \Delta T \) is the change in temperature (in °C) Given data: - Mass of water, \( m \) = 20.0 g - Initial temperature, \( T_i \) = 25.1°C - Final temperature, \( T_f \) = 35.1°C - Specific heat capacity of water, \( c \) = 4.184 J/g×°C The temperature change, \( \Delta T = T_f - T_i = 35.1°C - 25.1°C = 10.0°C \). Now substitute these values into the formula: \[ q = 20.0 \text{ g} \times 4.184 \text{ J/g} \times 10.0 \text{°C} \] \[ q = 837 J \] Therefore, the quantity of heat required is **837 Joules**. This example illustrates the practical application of the specific heat concept and demonstrates how to perform necessary calculations.
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