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lisabon 2012 [21]
3 years ago
15

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 fro

m 45.0°C to the boiling point and then boil away 0.700 kg of water?
Physics
1 answer:
hram777 [196]3 years ago
8 0

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

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Answer:

The deceleration is  a =  - 76.27 m/s^2

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