# How do you determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293 to 313 K?

Specific heat capacity of a substance is the amount of heat required to raise one gram of the substance by one degree Celsius. This means the first thing you need to do is convert your temperatures to Celsius.

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To determine the specific heat ((C)) of a material, you can use the formula:

[ Q = m \cdot C \cdot \Delta T ]

where:

- ( Q ) is the heat absorbed (in joules),
- ( m ) is the mass of the material (in grams),
- ( C ) is the specific heat of the material (in J/g·K),
- ( \Delta T ) is the change in temperature (in Kelvin).

Rearrange the formula to solve for ( C ):

[ C = \frac{Q}{m \cdot \Delta T} ]

Substitute the given values:

[ C = \frac{96 , \text{J}}{35 , \text{g} \cdot (313 , \text{K} - 293 , \text{K})} ]

[ C \approx \frac{96 , \text{J}}{35 , \text{g} \cdot 20 , \text{K}} ]

[ C \approx \frac{96 , \text{J}}{700 , \text{g} \cdot \text{K}} ]

[ C \approx 0.137 , \text{J/g} \cdot \text{K} ]

Therefore, the specific heat of the material is approximately (0.137 , \text{J/g} \cdot \text{K}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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