## Introductory Chemistry (5th Edition)

Therefore the final temperature of the aluminum is 32.8$^{\circ}$C.
We will use the equation $q=mc\Delta t$ q = heat transferred m = mass c = specific heat $\Delta t$ = change in temperature plug in the values. 67.4 = (265.95) (0.9) ($\Delta t$) ***We multiply 265.95 times 0.9 and then divide 67.4 by that answer *** solve using a calculator $\Delta t = 0.2810 ^{\circ}$ Therefore the temperature change is $0.2810^{\circ}$ The initial temp is 32.5 so the final temp= initial temp + $\Delta t$ = 32.5 + 0.2810 = 32.8. Therefore the final temperature is 32.8$^{\circ}$C.