If 0.022 kg of ice at 0C is added to 0.450 kg of water at 80C, what is the final temperature. (Assume that there is no heat lost to the surroundings.)
sum of heats lost is zero.
-.022*Hf+.450*Cw(80-Tf)+.022(Cw)(0-Tf)=0
solve for Tf
To find the final temperature of the system when ice is added to water, we need to use the principle of energy conservation. The heat lost by the hot water is equal to the heat gained by the ice to reach a final equilibrium temperature.
First, we need to determine the amount of heat lost by the hot water. We can use the formula:
Q = mcΔT
where Q is the heat lost, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
Mass of water (m1) = 0.450 kg
Initial temperature of water (T1) = 80°C
Next, we need to determine the amount of heat gained by the ice. The formula remains the same:
Q = mcΔT
where Q is the heat gained, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature.
Given:
Mass of ice (m2) = 0.022 kg
Initial temperature of ice (T2) = 0°C
To calculate the heat gained by the ice, we need to convert the ice from its initial temperature (0°C) to the final temperature (Tf), which is unknown. So the change in temperature for the ice would be Tf - 0°C.
Now, according to the principle of energy conservation, the heat lost by the water is equal to the heat gained by the ice. Therefore, we can equate the two equations:
m1c(Tf - T1) = m2c(Tf - T2)
Plugging in the given values:
(0.450 kg)(4,186 J/kg°C)(Tf - 80°C) = (0.022 kg)(2,093 J/kg°C)(Tf - 0°C)
Now we can solve for Tf:
(0.450 kg)(4,186 J/kg°C)(Tf - 80°C) = (0.022 kg)(2,093 J/kg°C)(Tf - 0°C)
Multiply out the equation:
((0.450 kg)(4,186 J/kg°C)Tf - (0.450 kg)(4,186 J/kg°C)(80°C)) = ((0.022 kg)(2,093 J/kg°C)Tf - 0)
Simplify the equation:
(0.450 kg)(4,186 J/kg°C)Tf - (0.450 kg)(4,186 J/kg°C)(80°C) = (0.022 kg)(2,093 J/kg°C)Tf
Now, isolate Tf:
(0.450 kg)(4,186 J/kg°C)Tf - (0.022 kg)(2,093 J/kg°C)Tf = (0.450 kg)(4,186 J/kg°C)(80°C)
Combine like terms:
(0.450 kg)(4,186 J/kg°C - 2,093 J/kg°C)Tf = (0.450 kg)(4,186 J/kg°C)(80°C)
Simplify:
(0.450 kg)(2,093 J/kg°C)Tf = (0.450 kg)(4,186 J/kg°C)(80°C)
Divide both sides by (0.450 kg)(2,093 J/kg°C) to isolate Tf:
Tf = ((0.450 kg)(4,186 J/kg°C)(80°C)) / ((0.450 kg)(2,093 J/kg°C))
Simplify further:
Tf = (334,560 J) / (940.35 J)
Calculate Tf:
Tf ≈ 355.86°C
Therefore, the final temperature of the system (after the ice is added to water) is approximately 355.86°C.