Question

How much heat transfer (in kcal) is required to raise the temperature of a 0.850 kg aluminum pot containing 2.00 kg of water from 45.0°C to the boiling point and then boil away 0.700 kg of water?

Answer 1

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

$\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T +m_w L_v$

Here,

$m_{Al}$= Mass of Aluminum

$C_p$= Specific Heat of Aluminum

$C_w$= Specific Heat of Water

$m_w =$ Mass of water

$L_v =$Latent of Vaporization

Replacing,

$\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)$

Converting,

$\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})$

$\Delta Q = 440.409kCal$

Therefore is required 440.409kCal