A10.0-gram sample of h2o(l) at 23.0°c absorbs 209 joules of heat. what is the final temperature of the h2o(l) sample? (1) 5.0°c (3) 28.0°c(2) 18.0°c (4) 50.0°c

option (3) 28.0°C

Explanation:

1) Data:

m = 10.0 g

Ti = 23.0°C

Q = 209 J

Tf = ?

2) Data from literature (textbook or internet)

Cs = 1.00 cal/g°C = 4.18 J/g°C

3) Formula:

Q = m Cs ΔT

4) Solution:

Q = m Cs ΔT = m Cs (Tf - Ti) ⇒ Tf - Ti = Q / (m Cs)

⇒ Tf = Ti + Q / (m Cs) = 23.0°C + 209 J/g°C / (10.0g × 4.18 J/g°C)

Tf = 23.0°C + 5.00 °C = 28.0°C

there are two significant figures.

i think tornado but not 100%