When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.
We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.
[tex]Q = c \times m \times \Delta T = \frac{4200J}{kg.\° C} \times 0.50kg \times (-22\° C) \times \frac{1kJ}{1000J} = -46.2 kJ[/tex]
where,
When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.
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