Question

How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.0ºC to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W

Answer 1

Answer:

Part a)

$Q = 2.47 \times 10^6$

Part b)

t = 4950.3 s

Explanation:

As we know that heat required to raise the temperature of container and water in it is given as

$Q = m_1s_2\Delta T_1 + m_2s_2\Delta T_2$

here we know that

$m_1 = 0.750$

$s_1 = 900$

$m_2 = 2.50 kg$

$s_2 = 4186$

$\Delta T_1 = \Delta T_2 = 100 - 30 = 70^oC$

now we have

$Q_1 = 0.750(900)(70) + (2.5)(4186)(70) = 779800 J$

now heat require to boil the water

$Q = mL$

here

m = 0.750 kg

$L = 2.25 \times 10^6 J/kg$

now we have

$Q_2 = 0.750(2.25 \times 10^6) = 1.7 \times 10^6 J$

Now total heat required is given as

$Q = Q_1 + Q_2$

$Q = 779800 + 1.7 \times 10^6 = 2.47 \times 10^6 J$

Part b)

Time taken to heat the water is given as

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

here we know that

power = 500 W

now we have

$t = \frac{2.47 \times 10^6}{500} = 4950.3 s$