Question

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.

Answer 1

The time is 0.713 sec.

Step-by-step 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

`Q_(b)=mc\Delta T`

Put the value into the formula

`Q_(b)=1.1*0.46*(70-12)`

`Q_(b)=29.348\ kJ`

We need to calculate the heat gained by water

Using formula of heat

`Q_(w)=mc\Delta T`

Put the value into the formula

`Q_(w)=4*4.18*(70-12)`

`Q_(w)=969.76\ kJ`

We need to calculate the total heat

Using formula of heat

`Q=Q_(b)+Q_(w)`

Put the value into the formula

`Q=29.348+969.76`

`Q=999.108\ kJ`

We need to calculate the time

Using formula of time

`t=(Q)/(P)`

Put the value into the formula

`t=(999.108)/(1400)`

`t=0.713\ sec`

Hence, The time is 0.713 sec.