How would you calculate the specific beat of silver, given that 280 g of the metal absorbed 136 calories when changed from 22.13°C to 30.79°C?
The equation you'll need is the following: Rearrange the equation to isolate
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To calculate the specific heat of silver, you can use the formula:
[ q = mc\Delta T ]
Where:
- ( q ) = heat absorbed (in calories)
- ( m ) = mass of the substance (in grams)
- ( c ) = specific heat capacity (in calories/gram °C)
- ( \Delta T ) = change in temperature (in °C)
Rearrange the formula to solve for specific heat (( c )):
[ c = \frac{q}{m\Delta T} ]
Substitute the given values:
[ c = \frac{136 \text{ cal}}{280 \text{ g} \times (30.79°C - 22.13°C)} ]
Calculate:
[ c ≈ \frac{136 \text{ cal}}{280 \text{ g} \times 8.66°C} ≈ \frac{136 \text{ cal}}{2420.8 \text{ cal}} ]
[ c ≈ 0.0562 \text{ cal/g°C} ]
So, the specific heat of silver is approximately ( 0.0562 \text{ cal/g°C} ).
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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.
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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