What would be the final temperature if #3.31 * 10^3# joules of heat were added to 18.5 grams of water at 22.0°C?
I found:
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To find the final temperature, you can use the formula:
( q = mcΔT )
Where:
- ( q ) = heat energy (Joules)
- ( m ) = mass of the substance (grams)
- ( c ) = specific heat capacity of the substance (J/g°C)
- ( ΔT ) = change in temperature (°C)
Rearranging the formula to solve for ( ΔT ), we get:
( ΔT = \frac{q}{mc} )
Then, you can use the calculated ( ΔT ) to find the final temperature using the formula:
( T_f = T_i + ΔT )
Given:
- ( q = 3.31 \times 10^3 ) Joules
- ( m = 18.5 ) grams
- ( c = 4.18 ) J/g°C (specific heat capacity of water)
- ( T_i = 22.0°C )
Calculate ( ΔT ) using the first formula, then add ( ΔT ) to ( T_i ) to find ( T_f ).
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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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