Answer
Therefore the final temperature of the water sample is 48.6$^{\circ}$ C.
Work Step by Step
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.
$1.7 \times 10^{4}$ = (245) (4.18) ($\Delta t $)
***We multiply 245 times 4.18 and then divide $1.7 \times 10^{4}$ by that answer
*** solve using a calculator
$ \Delta t = 16.6 ^{\circ} $
Therefore the temperature change is $1.82^{\circ}$
The initial temp is 155 so the final temp= initial temp + $ \Delta t$
= 32+ 16.6 = 48.6
Therefore the final temperature is 48.6$^{\circ}$ C
