A 500 mL bottle of water, which is at 25 degrees C, is poured over 120 g of ice at -8 degrees C. What will be the final temperature of the water when all the ice has melted. Assume that container is insulated and does not change temperature.
http://www.jiskha.com/display.cgi?id=1264029244
Does c in the equation stand for temperature in celsius?
To find the final temperature of the water when all the ice has melted, we need to apply the principle of conservation of energy. The heat lost by the ice as it melts will be gained by the water.
The heat lost or gained can be calculated using the following formula:
Q = m * c * ΔT
Where:
- Q is the heat lost or gained (in Joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)
First, let's calculate the heat gained by the ice as it melts. We know the mass of the ice is 120 g, and its specific heat capacity is 2.09 J/g°C. The change in temperature is the difference between the melting point of ice (0°C) and the initial temperature (-8°C):
Q_ice = 120 g * 2.09 J/g°C * (0°C - (-8°C))
Now, let's calculate the heat lost by the water. We know the initial temperature of the water is 25°C, its mass is 500 g, and the specific heat capacity of water is 4.18 J/g°C. The change in temperature is the difference between the final temperature of the water and the initial temperature (T_f - 25°C):
Q_water = 500 g * 4.18 J/g°C * (T_f - 25°C)
Since the heat lost by the ice is equal to the heat gained by the water (assuming no heat is lost to the surroundings), we can set up the equation:
Q_ice = Q_water
120 g * 2.09 J/g°C * (0°C - (-8°C)) = 500 g * 4.18 J/g°C * (T_f - 25°C)
Now we can solve for T_f:
120 g * 2.09 J/g°C * 8°C = 500 g * 4.18 J/g°C * (T_f - 25°C)
1996.8 J = 2090 J * (T_f - 25°C)
Divide both sides by 2090 J:
1996.8 J / 2090 J = T_f - 25°C
0.956 = T_f - 25°C
Add 25°C to both sides:
0.956 + 25°C = T_f
T_f ≈ 25.956°C
Therefore, the final temperature of the water when all the ice has melted will be approximately 25.956°C.