Answer:
45.70 C
Explanation:
We'll use the formula for specific heat Q=mcΔT. Let's first check out units and look at the values we have.
Q=heat= 203.93 J
m=mass= 9.02g
c= specific heat= 4.184 J/g K
ΔT= Change in temperature= Final temp. - initial temp. = [tex]T_{f}[/tex] - [tex]T_{i}[/tex]
Let's plug in what we know.
203.93 J = (9.02g)(4.184)( [tex]T_{f}[/tex] - 40.3 C)
We'll just solve this algebraically to solve for [tex]T_{f}[/tex]. It's like solving for x, just a different variable. It's easier to do the more complicated side first and then start moving your numbers to the other side.
203.93 = (37.73)( [tex]T_{f}[/tex] - 40.3) We'll multiply 37.73 by [tex]T_{f}[/tex] AND 40.3 so we'll get...
203.93 = (37.73Tf - 1520.5)
Let's start moving numbers to the other side
203.93 + 1520.5 = 37.73Tf
1724.4 = 37.73Tf
[tex]\frac{1724.4}{37.73}[/tex]= [tex]T_{f}[/tex]
45.70 C= [tex]T_{f}[/tex]
This makes sense with our problem since they heated water, so we should see an increase in temperature.
