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Sagot :
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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