22.3 is the answer to this question
<h2>Answer:</h2>
He is right that the energy of vaporization of 47 g of water s 106222 j.
<h3>Explanation:</h3>
Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.
In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.
It means for vaporizing 18 g, 40.65 kJ energy is needed.
So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ
Hence the student is right about the energy of vaporization of 47 g of water.
Answer:
87.3 calories of heat is required.
Explanation:
Heat = mcΔT
m= mass, c = specific heat of silver, T = temperature
H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)
H = 57.8 * 0.057 * 26.5
H = 87.3069 cal.
The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.
Answer:
yh
Explanation:
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<u>Answer:</u> The specific heat of metal is 0.821 J/g°C
<u>Explanation:</u>
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of metal = 30 g
= mass of water = 100 g
= final temperature = 25°C
= initial temperature of metal = 110°C
= initial temperature of water = 20.0°C
= specific heat of metal = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]](https://tex.z-dn.net/?f=30%5Ctimes%20c_1%5Ctimes%20%2825-110%29%3D-%5B100%5Ctimes%204.186%5Ctimes%20%2825-20%29%5D)
Hence, the specific heat of metal is 0.821 J/g°C
