kking38
Answered

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A 1250.0 g piece of aluminum at -35.0°C receives 72.41 kJ of heat. If aluminum has a heat capacity of 0.90
calculate AT in °C. The melting point of Al is 660°C. Your answer must have the correct number of sig figs.
g.K
AT =
°C

Sagot :

sebassandin

This problem is providing the mass, specific heat and melting point of aluminum and the amount of heat it receives in a heating process so it asks for the change in its temperature, which results in 64.36 °C.

Calorimetry

In chemistry, one can perform calorimetry calculations with the widely-known heat equation relating mass, specific heat, temperature and heat:

[tex]Q=mC\Delta T[/tex]

Thus, when having 1250.0 g of aluminum at -35.0 °C and supplying 72.41 kJ (72,410 J) of heat, we will be able to solve for the change in temperature by solving for it in the previous formula:

[tex]\Delta T=\frac{Q}{mC} \\\\\Delta T=\frac{72,410J}{1250.0g*0.90\frac{J}{g\°C} } \\\\\Delta T=64.36\°C[/tex]

Which is not as high as for it to provoke a melting change, with the correct significant figures as 0.90 do not contribute to this for being a theoretical number.

Learn more about calorimetry: brainly.com/question/1407669

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