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Sagot :
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: 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
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