Suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20 c. what would have been the final temperature of the water

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

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