Question

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

temperature of aluminum is 157.3°C, what is the final temperature? The specific heat
capacity of aluminum is 0.900 J/gºC.

Answer 1

`T_2= -90.0°C`

Step-by-step 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+(Q)/(mC)\\\\T_2=157.3\°C+(-6120.0J)/(27.5g*0.900 J/g\ºC)\\\\T_2= -90.0°C`

Regards!