Calcula cuanto costara tener encendido toda la noche (8horas) un radiador de 2500w sabiendo que el precio del Kwh es de 15 centimos

300 centavos

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Potencia (P) = 2500 W

Tiempo (t) = 8 horas

Costo de operación =?

A continuación, convertiremos 2500 W a KW. Esto se puede obtener de la siguiente manera:

1000 W = 1 KW

Por lo tanto,

2500 W = 2500 W × 1 KW / 1000 W

2500 W = 2,5 KW

Por tanto, 2500 W equivalen a 2,5 KW

A continuación, determinaremos la energía consumida. Esto se puede obtener de la siguiente manera: Potencia (P) = 2,5 KW

Tiempo (t) = 8 horas

Energía (E) =?

E = Pt

E = 2,5 × 8

E = 20 KWh

Finalmente, determinaremos el costo de operación. Esto se puede obtener de la siguiente manera:

1 KWh = 15 centavos

Por lo tanto,

20 KWh = 20 KWh × 15 centavos / 1 KWh

20 KWh = 300 centavos

Por lo tanto, costará 300 centavos tener un radiador de 2500 W encendido toda la noche (8 horas).

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.