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how many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?

Specific Heat Capacity of Aluminum: 0.897 J/g C

Sagot :

Sarah06109

Answer:

[tex]\boxed {\boxed {\sf 10.2 \ kJ}}[/tex]

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

[tex]q=mc \Delta T[/tex]

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

[tex]q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)[/tex]

Multiply the first two numbers. The units of grams cancel.

[tex]q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)[/tex]

[tex]q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)[/tex]

[tex]q= (201.825\ J / \textdegree C)(50.5 \textdegree C)[/tex]

Multiply again. This time, the units of degrees Celsius cancel.

[tex]q= 201.825 \ J * 50.5[/tex]

[tex]q= 10192.1625 \ J[/tex]

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

[tex]\frac { 1 \ kJ}{ 1000 \ J}[/tex]

Multiply by the answer we found in Joules.

[tex]10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}[/tex]

[tex]10192.1625 * \frac{ 1 \ kJ}{ 1000 }[/tex]

[tex]\frac {10192. 1625}{1000} \ kJ[/tex]

[tex]10.1921625 \ kJ[/tex]

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

[tex]10.2 \ kJ[/tex]

Approximately 10.2 kilojoules of energy would be required.

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