What would be the final temperature of the mixture when .5 g of ice at -10 degree are mixed with 250g of water at 30 degree centigrade? S.H.C of ice 0.5cal/g degree centigrade and S.L.H of ice = 80cal/g.
Same as last problem
mc change T of ice from -5 to 0
mL solid ice to liquid ice
mc change T melted ice to Tf
mc change T of water from 30 to Tf
To find the final temperature of the mixture, we can use the principle of conservation of energy.
First, let's calculate the heat absorbed by the ice to reach the final temperature. The heat absorbed by a substance can be calculated using the formula:
Q = m * s * ΔT
Where:
Q = heat absorbed/released (in calories)
m = mass of the substance (in grams)
s = specific heat capacity of the substance (in cal/g°C)
ΔT = change in temperature (in °C)
For the ice, the mass (m) is 0.5 g, specific heat capacity (s) is 0.5 cal/g°C, and the change in temperature (ΔT) is the final temperature minus the initial temperature (final temperature - initial temperature).
The initial temperature of the ice is -10°C, and we're trying to find the final temperature when it reaches thermal equilibrium with the water.
Let's denote the final temperature as T (°C). The change in temperature for the ice is T - (-10).
So, the heat absorbed by the ice is:
Q_ice = 0.5 g * 0.5 cal/g°C * (T - (-10))
Next, let's calculate the heat released by the water to reach the final temperature. The heat released by the water can also be calculated using the same formula:
Q = m * s * ΔT
For the water, the mass (m) is 250 g, specific heat capacity (s) is 1 cal/g°C (assuming it's liquid water), and the change in temperature (ΔT) is the final temperature minus the initial temperature (final temperature - initial temperature).
The initial temperature of the water is 30°C, and we need to find the final temperature.
So, the heat released by the water is:
Q_water = 250 g * 1 cal/g°C * (T - 30)
According to the principle of conservation of energy, the heat absorbed by the ice is equal to the heat released by the water:
Q_ice = Q_water
Therefore, we can equate the two quantities:
0.5 g * 0.5 cal/g°C * (T - (-10)) = 250 g * 1 cal/g°C * (T - 30)
Simplifying the equation:
0.25(T + 10) = 250(T - 30)
0.25T + 2.5 = 250T - 7500
249.75T = 7502.5
T ≈ 30.04°C
Therefore, the final temperature of the mixture would be approximately 30.04°C.