Question

A 10.0-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

Answer 1

The specific heat capacity of water is 4200J/(kg*℃). So when absorbs 209 joules, the water sample will increase 209/(4200*0.01)=5℃. So the final temperature of sample is 23+5=28℃.

Answer 2

Answer: option (3) 28.0°C

Step-by-step 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