The amount of heat (in calories) required to boil 2.8 Kg of water is 1512000 calories.
- We'll begin by converting 2.8 Kg of water to grams. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
2.8 Kg = 2.8 × 1000
2.8 Kg = 2800 g
- Finally, we shall determine the heat required to boil 2800 g (i.e 2.8 Kg) of water. This can be obtained as follow:
Mass (m) = 2800 g
Heat of Vaporisation (ΔHv) = 540 cal/g
Heat (Q ) = ?
Q = m•ΔHv
Q = 2800 × 540
Q = 1512000 calories
Thus, 1512000 calories is required to boil 2.8 Kg of water
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