What is the mass in grams of an object with a specific heat of 0.63 J/g . oC that absorbs 4,322 J of heat when the temperature changes from 301 oC to 664oC?

(write your answer to 3 decimal places, use zeros as place holders as needed)

To find the mass of the object, we can use the formula for heat absorbed:

\[
q = mc\Delta T
\]

Where:
- \( q \) is the heat absorbed (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).

We are given:
- \( q = 4322 \, \text{J} \),
- \( c = 0.63 \, \text{J/g°C} \),
- Initial temperature = 301°C,
- Final temperature = 664°C.

First, calculate the temperature change:

\[
\Delta T = 664°C - 301°C = 363°C
\]

Now, rearrange the formula to solve for the mass \( m \):

\[
m = \frac{q}{c \Delta T}
\]

Substitute the known values:

\[
m = \frac{4322 \, \text{J}}{0.63 \, \text{J/g°C} \times 363 \, \text{°C}}
\]

Now, let's calculate the mass.

\[
m = \frac{4322}{0.63 \times 363} = \frac{4322}{228.69}
\]

\[
m \approx 18.895 \, \text{g}
\]

Thus, the mass of the object is approximately **18.895 grams**.

Would you like further details or have any other questions?

Here are 5 related questions to explore:
1. How do you convert between calories and joules?
2. What is the specific heat capacity of water in comparison to the given object?
3. How would the mass change if the temperature difference increased?
4. How is the specific heat of a substance measured experimentally?
5. What role does specific heat play in thermal energy storage?

**Tip**: Specific heat is a material property that indicates how much heat energy is required to raise the temperature of a substance by 1°C per gram.

What is the mass in grams of an object with a specific heat of 0.63 J/g·°C that absorbs 4,322 J of heat when the temperature changes from 301°C to 664°C? (write your answer to 3 decimal places, use zeros as placeholders as needed)

Find out how to calculate the mass of an object with a specific heat of 0.63 J/g·°C that absorbs 4,322 J of heat, with the temperature changing from 301°C to 664°C. This detailed solution walks through heat transfer equations, thermodynamic principles, and algebraic steps. Suitable for students in Grades 10-12.

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