Question

A 27.5g piece of aluminum sits in a room and cools. It loses 6120.0 J of heat. If the initial

Answer 1

Answer:

$T_2= -90.0°C$

Explanation:

Hello!

In this case, according to the given description of how the temperature changes for aluminum in agreement to the loss of heat of 6120.0 J, we can use the following equation:

$Q=mC\Delta T=mC(T_2-T_1)$

Thus, by knowing Q, m, C and the initial temperature, we are able to obtain:

$T_2=T_1+\frac{Q}{mC}\\\\T_2=157.3\°C+\frac{-6120.0J}{27.5g*0.900 J/g\ºC}\\\\T_2= -90.0°C$

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