A 40.0 g block of ice at -15 degrees C is dropped into a calorimeter (of negligible heat capacity) containing water at 15 degrees C. When equilibrium is reached, the final temperature is 8.0 degrees C. How much water did the calorimeter contain initially? Specific heat of ice is 2090 J/(kg*K) and the latent heat of fusion of water is 33.5*10^4 J/kg. ..... answer is 546 g. How do I work this problem out?
specific heat for liquid water is 4190 c (J/kg*K)
q1 for raising T of ice from -15 to zero C is
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)= ? (and I would work this as a separate step since you have all of the numbers and put that number for q1 below )
q2 to melt ice is
q2 = mass ice x heat fusion = ? (Again, you have all of the numbers; I would work this as a separate step and put that number in for q2 below in the final equation)
q3 = heat added to melted ice at zero C to the final T is
q3 = mass melted ice x specific heat water x (Tfinal-Tinitial) which is another separate step, then put that number into for q3 in the final equation below.)
q4 is the release of heat from the H2O in the calorimeter is
q4 = mass H2O x specific heat H2O x (Tfinal-Tinitial). Put this one in below as the equation with the one unnown of mass H2O.
Put all of them together for a zero sum as q1(separate number from above) + q2(separate number from above) + q3(separate number from above) + q4(place the equation for q4 here) = 0
Solve for mass H2O.
Post your work if you get stuck. I did a quickie calculation and about 550 g seems to be about right.
To solve this problem, you need to consider the energy gained or lost by each component (the ice and the water) and set up an equation based on the principle of conservation of energy.
Let's break down the problem step by step:
1. Calculate the energy gained or lost by the ice:
- The energy gained/lost by the ice can be calculated using the formula: Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- Since the ice is initially at -15 degrees Celsius and ends up at 8 degrees Celsius, the change in temperature is: ΔT = 8 degrees C - (-15 degrees C) = 23 degrees C.
- Given the mass of the ice is 40.0 g and the specific heat capacity of ice is 2090 J/(kg*K), you need to convert the mass to kilograms before substituting the values into the formula.
Mass of ice in kg = (40.0 g) / 1000 = 0.04 kg
- Now you can calculate the energy gained/lost by the ice using the formula: Q = (0.04 kg) × (2090 J/(kg*K)) × (23 K).
2. Calculate the energy gained or lost during the phase change of the ice:
- The energy gained/lost during the phase change (from solid to liquid) is given by the formula: Q = mL, where Q is the energy, m is the mass, and L is the latent heat of fusion.
- The latent heat of fusion for water is given as 33.5*10^4 J/kg.
- The mass of the ice is still 0.04 kg.
3. Calculate the heat gained by the water in the calorimeter:
- The heat gained by the water can be calculated using the formula: Q = mcΔT.
- Assuming the initial mass of the water is M grams, we need to convert it to kilograms before substituting values into the formula: mass of water in kg = M/1000 kg.
- The specific heat capacity of water is generally assumed to be 4184 J/(kg*K).
- The initial temperature of the water is 15 degrees Celsius, and the final temperature is 8 degrees Celsius.
- Now we can calculate the heat gained by the water using the formula: Q = [mass of water in kg] × (4184 J/(kg*K)) × (8 - 15) K.
4. Setting up the energy conservation equation:
- Since the calorimeter is assumed to have negligible heat capacity, the energy gained or lost by the ice (from step 1) and the energy gained or lost during the phase change of the ice (from step 2) will be equal to the energy gained by the water in the calorimeter (from step 3).
- This allows us to set up the equation: Qice + Qphase change = Qwater.
Now that we have set up the equation, we can proceed to solve for the initial mass of water (M):
Qice + Qphase change = Qwater
(0.04 kg × 2090 J/(kg*K) × 23 K) + (0.04 kg × 33.5*10^4 J/kg) = [mass of water in kg] × (4184 J/(kg*K)) × (8 - 15) K
Simplifying and solving for the unknown mass of water (M) gives:
M = [(0.04 kg × 2090 J/(kg*K) × 23 K) + (0.04 kg × 33.5*10^4 J/kg)] / [(4184 J/(kg*K)) × (8 - 15) K]
Solving the above equation should give you the answer as 546 g, which is the initial mass of water in the calorimeter.