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A small electric immersion heater is used to heat 97 g of water for a cup of instant coffee. The heater is labeled "63 watts" (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from 22°C to 100°C, ignoring any heat losses. (The specific heat of water is 4186 J/kg.K.)

Respuesta :

Answer:8.37 min

Explanation:

Given

mass of water(m)=97 gm

Heater labelled as  63 Watt

Change in temperature of water=100-22=[tex]78^{\circ}[/tex]

Heat required to raise temperature is

[tex]Q=mc\left ( \Delta T\right )[/tex]

[tex]Q=0.097\times 4186\times \left ( 78\right )[/tex]

Q=31,671.276 J

This energy is given by Heater

E=Pt

[tex]63\times t=31,671.276[/tex]

[tex]t=\frac{31,671.276}{63}[/tex]

t=502.71 s

t=8.37 min

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