## Introductory Chemistry (5th Edition)

Therefore the final temperature of the aluminum sample is 95.1$^{\circ}$ 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. $2.87 \times 10^{3}$ = (53.2) (0.9) ($\Delta t$) ***We multiply 53.2 times 0.9 and then divide $2.87 \times 10^{3}$ by that answer *** solve using a calculator $\Delta t = 59.9 ^{\circ}$ Therefore the temperature change is $1.82^{\circ}$ The initial temp is 155 so the final temp= initial temp - $\Delta t$ = 155 - 59.9 = 95.1 Therefore the final temperature is 95.1$^{\circ}$ C