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A small electric immersion heater is used to heat 68 g of water for a cup of instant coffee. The heater is labeled "120 watts" (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from 23°C to 100°C, ignoring any heat losses. (The specific heat of water is 4186 J/kg.K.)

Respuesta :

Answer:

t =  182.65 sec

Explanation:

Given Data:

[tex]T_1 =23 degree [/tex]

[tex]T_2 =100 degree [/tex]

mass of water = 68.0 gm

[tex]Q = mS ( T_2 - T_1)[/tex]

[tex]Q = 0.068\times 4180 (100 -23)[/tex]

Q = 21917.896 J

Time of heating = t

Heat generated Q = 21917.896 J

Heat power =[tex] p =\frac{Q}{t}[/tex]

[tex]t = \frac{Q}{120} = 182.65 sec[/tex]

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