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

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

$Q_{b}=mc\Delta T$

Put the value into the formula

$Q_{b}=1.1\times0.46\times(70-12)$

$Q_{b}=29.348\ kJ$

We need to calculate the heat gained by water

Using formula of heat

$Q_{w}=mc\Delta T$

Put the value into the formula

$Q_{w}=4\times4.18\times(70-12)$

$Q_{w}=969.76\ kJ$

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

$t=\dfrac{Q}{P}$

Put the value into the formula

$t=\dfrac{999.108}{1400}$

$t=0.713\ sec$

Hence, The time is 0.713 sec.