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A steel bucket contains 4 liters of water at 128C. An electric immersion heater rated at 1400 Watts is placed in the * Indicates an advanced level in solving. Problems 61 62 CHAPTER 1 Introduction bucket. Determine how long it will take for water to heat to 708C. Assume that the empty bucket weighs 1.1 kg. The specific heat of steel is 0.46 kJ/(kg 8C). Use an average specific heat of water of 4.18 kJ/(kg 8C). Disregard any heat loss to the surroundings.

Respuesta :

Answer:

The time is 0.713 sec.

Explanation:

Given that,

Weight of water = 4 liters

Initial temperature = 128°C

Power = 1400 Watts

Final temperature = 708°C

Weight = 1.1 kg

Specific heat of steel = 0.46 kJ/kg°C

Specific heat of water = 4.18 kJ/kg°C

We need to calculate the heat gained by bucket

Using formula of heat

[tex]Q_{b}=mc\Delta T[/tex]

Put the value into the formula

[tex]Q_{b}=1.1\times0.46\times(70-12)[/tex]

[tex]Q_{b}=29.348\ kJ[/tex]

We need to calculate the heat gained by water

Using formula of heat

[tex]Q_{w}=mc\Delta T[/tex]

Put the value into the formula

[tex]Q_{w}=4\times4.18\times(70-12)[/tex]

[tex]Q_{w}=969.76\ kJ[/tex]

We need to calculate the total heat

Using formula of heat

[tex]Q=Q_{b}+Q_{w}[/tex]

Put the value into the formula

[tex]Q=29.348+969.76[/tex]

[tex]Q=999.108\ kJ[/tex]

We need to calculate the time

Using formula of time

[tex]t=\dfrac{Q}{P}[/tex]

Put the value into the formula

[tex]t=\dfrac{999.108}{1400}[/tex]

[tex]t=0.713\ sec[/tex]

Hence, The time is 0.713 sec.

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