An ice cube has mass 9.0g and is added to a cup of coffee. The coffee's intial temp. is 90.0 deg. Celsius and the cup contains 120.0 g of liquid. Assume the specific heat capacity of coffee is the same as that of water. The heat of fusion of ice (associated with ice melting) is 6.0kJ/mol. Find the temp. of coffee after ice melts.
How do we get rid of the /mol under the kJ?
Also, may I get a hint please?
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
So water has molar mass 18 g/mol and you have an ice cube of 9.0 g, that looks like half a mole of ice.
The heat of fusion of ice is 6.0 kJ/mol and you have 0.5 moles of ice, so the energy used to melt your ice cube is 3.0 kJ
That answers the first part of your question, the next thing to do is to take the specific heat capacity of water and figure out how much those 3.0 kJ are going to change the overall temperature of the (129.0g) of coffee. Hope that helps.