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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.

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Determine the specific heat of the metal.

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  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

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