1 / 7

Determine the specific heat of the metal.

If 85.0 g of a metal at 90.0 o C is added to 100.0 g of water at 25.0 o C, the final temperature of the mixture is 30.9 o C. (The specific heat of water is 1.00 cal/g• o C.). Determine the specific heat of the metal. Remember: calories gained by water = calories lost by metal.

todd-benjamin
Download Presentation

Determine the specific heat of the metal.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. If 85.0 g of a metal at 90.0 oC is added to 100.0 g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•oC.) Determine the specific heat of the metal. Remember: calories gained by water = calories lost by metal

  2. If 85.0 g of a metal at 90.0 oC is added to 100.0 g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•K.) Determine the specific heat of the metal. calories gained by water = calories lost by metal 100.0 g 5.9 K H2O: 590 = cal Can we use  oC for  K?

  3. If 85.0 g of a metal at 90.0 oC is added to 100.0 g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•K.) Determine the specific heat of the metal. calories gained by water = calories lost by metal 100.0 g 5.9 K H2O: 590 = cal Now for the metal.

  4. If 85.0 g of a metal at 90.0 oC is added to 100.0 g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•K.) Determine the specific heat of the metal. calories gained by water = calories lost by metal 100.0 g 5.9 K H2O: 590 = cal 590 cal Sp.Ht. Metal 0.117 85.0 g 59.1 oC

  5. 85.0 g of a metal at 90.0 oC is added to 100.0 g of water at 25.0 oC. Determine the final temperature of the mixture. (The specific heat of water is 1.00 cal/g•K and the specific heat of the metal is 0.117 cal/g•K. ) calories gained by water = calories lost by metal (TF – TI) K 100.0 g 85.0 g (TI – TF)K = T must be positive

  6. 85.0 g of a metal at 90.0 oC is added to 100.0 g of water at 25.0 oC. Determine the final temperature of the mixture. (The specific heat of water is 1.00 cal/g•K and the specific heat of the metal is 0.117 cal/g•K. ) calories gained by water = calories lost by metal (TF – TI) K 100.0 g 85.0 g (TI – TF)K = T must be positive Insert the TI Values and Solve for TF TF = 30.0 oC

More Related