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?

Answer 1

I found: #T_f=64.6^@C#

We can try using the general relationship between heat #Q# (lost or gained) and change in temperature #DeltaT# of a mass #m# of a substance of specific heat #c# (in our case water: #c_("water")=4.2J/(g^@C)#):
#Q=mcDeltaT#
with our data: #3.31xx10^3=18.5*4.2*(T_f-22)# rearranging we get: #T_f=64.6^@C#
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Answer 2

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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Answer from HIX Tutor

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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