Answer:
The final temperature of water is 54.5 °C.
Explanation:
Given data:
Energy transferred = 65 Kj
Mass of water = 450 g
Initial temperature = T1 = 20 °C
Final temperature= T2 = ?
Solution:
First of all we will convert the heat in Kj to joule.
1 Kj = 1000 j
65× 1000 = 65000 j
specific heat of water is 4.186 J /g. °C
Formula:
q = m × c × ΔT
ΔT = T2 - T1
Now we will put the values in Formula.
65000 j = 450 g × 4.186 J /g. °C × (T2 - 20°C )
65000 j = 1883.7 j /°C × (T2 - 20°C )
65000 j/ 1883.7 j /°C = T2 - 20°C
34.51 °C = T2 - 20°C
34.51 °C + 20 °C = T2
T2 = 54.5 °C
