Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 3 years ago

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?

  • This Question is Closed
  1. benjaminf
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy