Question

Calculate the time required to heat 20 kg of water from 10°C to 35°C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 42000 J kg−1 K−1.

Answer 1

The time required to heat 20 kg of water from 10°C to 35°C using an immersion heater rated 1000 W is 44 min

Explanation:

Step 1:

Given data  

P = 1000 W

S = 4200 J kg−1 K−1

M  = 20 kg

ΔT = 25 °C

Where  

P is Power rating of the immersion rod,  

S is Specific heat of water,  

M is Mass of water,  

ΔT  is Change in temperature,

Step 2:

The total heat needed to elevate the temperature from 10 ° C to 35 ° C by 20 kg of water

Q = M × S × ΔT

Q = 20 × 4200 × 25

Q = 20 × 4200 × 25

   = $21 \times 10^{5} J$

Step  3:

Let the time it takes from 10 ° C to 35 ° C to heat 20 kg of water be t. When heating water, just 80 percent of the heat of the immersion rod is useful. Accordingly,

Immersion rod energy used to heat the water

= $t \times(0.80) \times 1000 J=21 \times 105 J$

$t=\frac{21 \times 10^{5}}{800}=2625 s$

$t=\frac{2625}{60}=43.75 \mathrm{min}$

≈44 min

Answer 2

The time required to heat 20 kg of water from 10°C to 35°C is 44 minutes.

Given:

Power = P = 1000 W

Mass = M = 20 Kg

Temperature change = ΔT = 35°C - 10°C = 25°C

Specific heat capacity of water = S = 42000 J/Kg.K

Explanation:

Q = M × S × ΔT

On substituting the values, we get,

Q = 20 × 4200 × 25

Q = 21 × 10⁵ J

From question, 80% of the power input is used to heat the water.

Thus, the energy utilized is given as:

Q = t × (0.8 × 1000)

21 × 10⁵ = t × (800)

t = (21 × 10⁵)/800

∴ t = 2625 s

Now, the time in minutes is:

t = 2625/60 = 43.75 min

∴ t ≈ 44 min