What is the final temperature of 400 g of water at 20°C after it absorbs 226 kJ of heat?
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To find the final temperature, use the equation ( q = mc\Delta T ) where ( q ) is the heat absorbed, ( m ) is the mass of the substance, ( c ) is the specific heat capacity, and ( \Delta T ) is the change in temperature. Rearrange the equation to solve for ( \Delta T ), then add ( \Delta T ) to the initial temperature to find the final temperature. Given:
- Mass (( m )) = 400 g
- Initial temperature (( T_i )) = 20°C
- Heat absorbed (( q )) = 226 kJ
- Specific heat capacity of water (( c )) = 4.18 J/g°C
( q = mc\Delta T ) ( \Delta T = \frac{q}{mc} )
( \Delta T = \frac{226000 \text{ J}}{400 \text{ g} \times 4.18 \text{ J/g°C}} ) ( \Delta T \approx 135.88 \text{°C} )
Final temperature (( T_f )) = Initial temperature (( T_i )) + ( \Delta T ) ( T_f = 20°C + 135.88°C ) ( T_f \approx 155.88°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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