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.

e% = 3.4%

Explanation:

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

         m (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%

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

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