Question

A heater is used to heat water in a bucket at an energy rate (power) of 4151 J/s. The bucket is made of aluminum and weigh's 1.2 kg. Calculate how long will it take to heat 20 kg of water from 20°C to 80°C? Express your answer in whole units shown.
Hint & useful data:
Assume that all heat from the heater "goes" to the water and bucket. The water and bucket are in thermal equilibrium at all times.
cw = 4186 J/kg°C
cal = 900J/kg°C

Please show working thank you :)

Answer 1

Time required to heat water = 20.18 min.

Given:

A heater is used to heat water in a bucket at an energy rate (power) of

4151 J/s.

The bucket is made of aluminum and weigh's 1.2 kg.

20 kg of water from 20°C to 80°C.

$C_w$ = 4186 J/kg°C

$C_{al}$ = 900J/kg°C

To find:

Time required to heat water.

Explanation:

First we find change in temperature of aluminium per second if we heat at a rate of 4151 J/s.

m$C_{al}$${\Delta T}_{al}$ = 4151 J/s.

        ${\Delta T}_{al}$= $\frac{4151}{1.2\times 900}$ = 3.843 Per second.

Now we can calculate the change in temperature of Water for  ${\Delta T}_{al}$ =3.843 Per second.

From energy balance equation.

m$C_{al}$${\Delta T}_{al}$ =m$C_{w}$${\Delta T}_{w}$

       ${\Delta T}_{w}$ = $\frac{1.2\times 900 \times 3.84}{20\times 4186}$ = 0.0495 per second.

From 20°C to 80°C we can increase the temperature by 60°c

So time required for increase the temperature by 60°c = 60 ÷ 0.0495

                                                                                            = 1211.2 sec.

                                                                                           = 20.18 min.

Time required to heat water = 20.18 min.

To learn more...

1)An electric heater takes 6A current from a 230V supply line,calculate the power of the heater and electric energy consumed by it in 5 hours.​

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