Question

Read the intro to lab 10. We are going to assume that the experiment was done in a styrofoam cup, so we are not going to include the heat gained by the cup or the stirrer, A piece of hot metal is put into cool water. Conservation of energy requires that heat lost = heat gained. So, m c delta T for water = m c delta T for the metal. c for water is 4,19 j/gC Data: mass of water = 100 g starting temp of water = 20 C mass of metal = 60 g starting temp of metal = 100 C Final temp of water and metal = 24 C Calculate the specific heat of the metal, determine what metal it might be from the list in the lab manual and calculate a percent error. Enter the result of your calculations in the quiz named specific heat. It will count as a lab.

Answer 1

Answer:

e% = 3.4%

Explanation:

This is a calorimetry problem where the heat released equals the heat absorbed

         m $c_{e1}$ (T₀ - T_f) = M c_{e2} (T₁ - T_f)

Index 1 refers to water and index 2 to metal, in this case it asks for the specific heat of the metal (c_{e2})

        c_{e2} = m / M c_{e1} (T_f -T₀) / (T₁ - T_f)

Let's calculate

      c_{e} = 60/100 4.19 (24-20) / (100-24)

      c_{e2} = 0.1323 j / gC

This metal is possibly lead, which is its specific heat is 0.128 J / gC

The percentage error is

        e% = (c_{e2} - 0.128) /0.128   100

         e% = 3.4%