An ice cube at its freezing point is placed in nested Styrafoam cups which contain 240 g of water at 21.5 °C. If the final equilibrium temperature when the last trace of ice melts is 25.0 °C, what was the mass of the ice cube?
The sum of heats gained is zero.
massice=m
m*Hf+m*c*(25-0)+240*C*(25-21.5)=0
solve for m
To find the mass of the ice cube, we can use the principle of conservation of energy. The heat lost by the water is equal to the heat gained by the ice cube.
First, let's calculate the heat lost by the water using the formula:
Q = m * c * ΔT
Where:
Q is the heat lost or gained
m is the mass of the substance (water in this case)
c is the specific heat capacity of the substance
ΔT is the change in temperature
The specific heat capacity of water is approximately 4.18 J/g°C.
Q_water = 240 g * 4.18 J/g°C * (25.0 °C - 21.5 °C)
Simplifying the equation, we get:
Q_water = 240 g * 4.18 J/g°C * 3.5 °C
Q_water = 3528 J
Since the heat lost by the water is equal to the heat gained by the ice cube, we can calculate the mass of the ice cube using the formula:
Q_ice = m_ice * ΔH_fusion
Where:
Q_ice is the heat gained by the ice cube
m_ice is the mass of the ice cube
ΔH_fusion is the heat of fusion of ice, which is 334 J/g
Q_ice = m_ice * ΔH_fusion
3528 J = m_ice * 334 J/g
Simplifying the equation, we get:
m_ice = 3528 J / 334 J/g
m_ice ≈ 10.57 g
Therefore, the mass of the ice cube is approximately 10.57 grams.