Answer :
To find the specific heat of the substance, we can use the formula:
[tex]\[ q = m C_0 \Delta T \][/tex]
where:
- \( q \) is the heat added (in joules),
- \( m \) is the mass (in grams),
- \( C_0 \) is the specific heat (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).
Given data:
- Mass (\( m \)) = 10.0 kg
- Initial temperature (\( T_{initial} \)) = 10.0°C
- Final temperature (\( T_{final} \)) = 70.0°C
- Heat added (\( q \)) = 2,520 J
Let's solve the problem step by step:
1. Convert mass from kilograms to grams:
[tex]\[ m = 10.0 \, \text{kg} \times 1000 \, \frac{\text{g}}{\text{kg}} = 10000 \, \text{g} \][/tex]
2. Calculate the change in temperature:
[tex]\[ \Delta T = T_{final} - T_{initial} = 70.0^{\circ}\text{C} - 10.0^{\circ}\text{C} = 60.0^{\circ}\text{C} \][/tex]
3. Rearrange the formula to solve for specific heat (\( C_0 \)):
[tex]\[ C_0 = \frac{q}{m \Delta T} \][/tex]
4. Substitute the given values into the formula:
[tex]\[ C_0 = \frac{2,520 \, \text{J}}{10000 \, \text{g} \times 60.0^{\circ}\text{C}} \][/tex]
5. Calculate the specific heat:
[tex]\[ C_0 = \frac{2,520}{600,000} \, \text{J/g°C} \][/tex]
[tex]\[ C_0 = 0.00420 \, \text{J/g°C} \][/tex]
The specific heat \( C_0 \) of the substance is:
[tex]\[ 0.00420 \, \text{J/g°C} \][/tex]
Among the given options, the correct one is:
[tex]\[ \boxed{0.00420 \, \text{J/g°C}} \][/tex]
[tex]\[ q = m C_0 \Delta T \][/tex]
where:
- \( q \) is the heat added (in joules),
- \( m \) is the mass (in grams),
- \( C_0 \) is the specific heat (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).
Given data:
- Mass (\( m \)) = 10.0 kg
- Initial temperature (\( T_{initial} \)) = 10.0°C
- Final temperature (\( T_{final} \)) = 70.0°C
- Heat added (\( q \)) = 2,520 J
Let's solve the problem step by step:
1. Convert mass from kilograms to grams:
[tex]\[ m = 10.0 \, \text{kg} \times 1000 \, \frac{\text{g}}{\text{kg}} = 10000 \, \text{g} \][/tex]
2. Calculate the change in temperature:
[tex]\[ \Delta T = T_{final} - T_{initial} = 70.0^{\circ}\text{C} - 10.0^{\circ}\text{C} = 60.0^{\circ}\text{C} \][/tex]
3. Rearrange the formula to solve for specific heat (\( C_0 \)):
[tex]\[ C_0 = \frac{q}{m \Delta T} \][/tex]
4. Substitute the given values into the formula:
[tex]\[ C_0 = \frac{2,520 \, \text{J}}{10000 \, \text{g} \times 60.0^{\circ}\text{C}} \][/tex]
5. Calculate the specific heat:
[tex]\[ C_0 = \frac{2,520}{600,000} \, \text{J/g°C} \][/tex]
[tex]\[ C_0 = 0.00420 \, \text{J/g°C} \][/tex]
The specific heat \( C_0 \) of the substance is:
[tex]\[ 0.00420 \, \text{J/g°C} \][/tex]
Among the given options, the correct one is:
[tex]\[ \boxed{0.00420 \, \text{J/g°C}} \][/tex]