Question

A steel bucket contains 4 liters of water at 12 °C. An electric immersion heater rated at 1400 Watts is placed in the bucket. Determine how long it will take for water to heat to 70 °C. Assume that the empty bucket weighs 1.1 kg. The specific heat of steel is 0.46 kJ/(kg °C). Use an average specific heat of water of 4.18 kJ/(kg °C). Disregard any heat loss to the surrounding

Answer 1

Answer:

It will take 11.9 minutes to heat the water to 70 degree C

Explanation:

As we know that thermal energy gain from electric heater is used to increase the temperature of bucket and water both

So we will first find the energy required to increase the temperature of Water and Bucket

$Q = m_1 c_1\Delta T + m_2c_2\Delta T$

$Q_1 = 4(4.18)(70 - 12) + 1.1(0.46)(70 - 12)$

$Q_1 = 999.11 kJ$

Now we know that heat given by the heater is

$Q = P . t$

$P = 1400 Watt$

now we have

$1400 t = 999.11 \times 10^3$

$t = 713.6 s$

$t = 11.9 minutes$

#Learn

Topic : Thermal Energy Transfer

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