The heat is added to 200g ice at 0ºC to make the ice become water. there is no increase in temperature. the heat added is equal to the number of moles of ice melted x molar heat of fusion of ice, 6.01kJ/mole.
<span>200g / 18g/mole = 11.11moles ice </span>
<span>11.11moles x 6.01kJ/mole = 66.78 kJ heat just to melt the ice to water </span>
<span>if you are just looking for the heat to melt the sample of ice, this is where you stop. </span>
<span>if you need to determine the heat to melt the ice AND raise the temperature, this is what you require </span>
<span>the second phase is raising the temperature of the water to 65ºC. this is equal to the mass of the water x specific heat of water x change in temperature </span>
<span>200g x 4.187J/g-ºC x 65ºC = 54431J or 54.431kJ energy to raise the water from 0ºC to 65ºC </span>
<span>the total heat required to melt the ice and then raise the temperature to 65ºC: </span>
<span>66.78kJ + 54.431kJ = 121.21kJ heat</span>
Answer:
(4) 230 kPa
Explanation:
The temperature is constant, so the only variables are pressure and volume.
We can use Boyle’s Law.
Divide both sides of the equation by 
.
Because a unexpected color change has taken place also a precipitate was formed
Same, I cannot answer the question. Try asking a high school student
