A piece of metal at a temperature of 100 °C is dropped into an equal mass of water at atemperature of 15 °C in a container of negligible mass. The specific heat capacity of wateris four times that of the metal. What is the final temperature of the mixture
Question
A piece of metal at a temperature of 100 °C is dropped into an equal mass of water at atemperature of 15 °C in a container of negligible mass. The specific heat capacity of wateris four times that of the metal. What is the final temperature of the mixture
Solution
To solve this problem, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In this case, the heat lost by the metal will be equal to the heat gained by the water.
Let's denote:
- The final temperature of the mixture as T (°C)
- The specific heat capacity of the metal as c (J/g°C)
- The specific heat capacity of water as 4c (since it's four times that of the metal)
- The mass of the metal and water as m (g)
The heat lost by the metal is given by Q_metal = mc(T_initial_metal - T) and the heat gained by the water is Q_water = 4mc(T - T_initial_water).
Since the heat lost by the metal is equal to the heat gained by the water, we can set these two equations equal to each other:
mc(T_initial_metal - T) = 4mc(T - T_initial_water)
Solving for T gives:
T = (mcT_initial_metal + 4mcT_initial_water) / (5mc)
Substituting the given values:
T = (mc100 + 4mc15) / (5m*c)
The m's and c's cancel out:
T = (100 + 4*15) / 5 = 70°C
So, the final temperature of the mixture is 70°C.
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