if 5.25g of ice AT 0 c IS PLACED IN 25.63 G OF LIQUID WATER AT 28c, WHAT WILL BE THE FINAL TEMPERATURE OF THIS MIXTURE?
To find the final temperature of the mixture, we can use the principle of conservation of energy, which states that the total heat gained by the ice and the heat lost by the water will be equal at thermal equilibrium.
The heat gained by the ice can be calculated using the equation:
Q_ice = mass_ice * specific_heat_ice * change_in_temperature
The heat lost by the water can be calculated using the equation:
Q_water = mass_water * specific_heat_water * change_in_temperature
At thermal equilibrium, Q_ice = Q_water.
Since the ice is at 0°C and the water is at 28°C, we have:
Q_ice = 5.25g * specific_heat_ice * (final_temperature - 0°C)
Q_water = 25.63g * specific_heat_water * (28°C - final_temperature)
Setting the two equations equal to each other, we get:
5.25g * specific_heat_ice * (final_temperature - 0°C) = 25.63g * specific_heat_water * (28°C - final_temperature)
Simplifying, we have:
5.25 * specific_heat_ice * final_temperature = 25.63 * specific_heat_water * (28°C - final_temperature)
Now we can solve for the final temperature.
Note: The specific heat capacity of ice is 2.09 J/g°C, and the specific heat capacity of water is 4.18 J/g°C.
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