An ice cube at 0 degrees Celsius was dropped into 30.0g of water in a cup at 45 degrees Celsius. At the instant that all of the ice was melted, the temperature of the water in the cup was 19.5 degrees Celsius. What was the mass of the ice cube?
How much heat did the water lose in going from 45 to 19.5 degrees?
Mass x specific heat water x (Tfinal-Tinitial).
How much heat did the ice gain?
mass ice x heat fusion.
heat lost + heat gain = 0.
The only unknown is mass ice.
I know the answer is 9.59 grams somehow but I can only manage to get 79.104 g
To solve this problem, we need to use the principle of conservation of energy.
Step 1: Calculate the heat absorbed by the water to reach 19.5 degrees Celsius.
The heat absorbed by the water can be calculated using the equation:
Q = mcΔT
where:
Q = heat absorbed by the water (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (which is 4.18 J/g·°C)
ΔT = change in temperature of the water (from 45°C to 19.5°C)
Given that the mass of the water (m) is 30.0g, ΔT = (19.5°C - 45°C), and c = 4.18 J/g·°C, we can calculate the heat absorbed by the water:
Q = (30.0g) * (4.18 J/g·°C) * (19.5°C - 45°C)
Step 2: Calculate the heat released by the ice cube to reach 0°C.
The heat released by the ice cube can be calculated using the equation:
Q = mL
where:
Q = heat released by the ice cube (in Joules)
m = mass of the ice cube (in grams)
L = heat of fusion of water (which is 334 J/g)
Since the ice is melting and its temperature is changing from 0°C to 0°C, the change in temperature (ΔT) is 0°C. Therefore, the heat released by the ice cube is:
Q = (m) * (334 J/g)
Step 3: Equate the heat absorbed by the water to the heat released by the ice cube.
Since the ice cube is melting into the water, the heat absorbed by the water is equal to the heat released by the ice cube:
(30.0g) * (4.18 J/g·°C) * (19.5°C - 45°C) = (m) * (334 J/g)
Step 4: Solve for the mass of the ice cube (m).
We can rearrange the equation to solve for m:
m = (30.0g) * (4.18 J/g·°C) * (19.5°C - 45°C) / (334 J/g)
m ≈ -16.779g (rounded to three decimal places)
Since mass cannot be negative, the mass of the ice cube must be zero. This means that the initial water at 45 degrees Celsius was completely cooled down to 19.5 degrees Celsius without the addition of any ice cube.
To find the mass of the ice cube, we can use the principle of conservation of energy. The energy gained by the ice cube when it melts equals the energy lost by the water when it cools down.
First, we need to find the energy gained by the ice cube when it melts. We can use the formula:
Q = mcΔT
Where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the ice cube is at 0 degrees Celsius and we want to find the energy gained when it melts, we need to use the heat of fusion (latent heat) for water, which is 334 J/g.
The mass of the ice cube is denoted as m_ice.
The energy gained by the ice cube when it melts is given by:
Q_ice = m_ice * 334 J/g
Next, we need to calculate the energy lost by the water. We use the same formula, but this time the change in temperature is from 45 degrees Celsius to 19.5 degrees Celsius:
Q_water = 30.0g * 4.18 J/g °C * (45 - 19.5) °C
Now, according to the principle of conservation of energy, the energy gained by the ice cube is equal to the energy lost by the water:
Q_ice = Q_water
m_ice * 334 J/g = 30.0g * 4.18 J/g °C * (45 - 19.5) °C
Simplifying the equation, we can solve for m_ice:
m_ice = (30.0g * 4.18 J/g °C * (45 - 19.5) °C) / (334 J/g)
Calculating this expression, we find that the mass of the ice cube is approximately 26.7 grams.