An ice cube at 0 degrees Celsius is placed in a pouch and dropped into 30.0g of water in a cup at 45 degrees Celsius. The ice cube melts and the resulting water warms up. The lowest temperature reached by the 30.0g of water in the cup was 20.0 degrees Celsius. What was the mass of the ice cube that was placed in the pouch?
- Please help me, I don't even know where to start or if where I think I have to start is even correct. Thank you.
Why didn't you tell us where you THINK you should start?
Let y = mass ice cube
To melt ice cube at zero C it is y*heat fusion. Add that to in as this,
Then (y*heat fusion) + [y*specific heat H2O x (Tfinal-Tinitial)] + [(mass H2O in cup x specific heat H2O x (Tfina-Tinitial)] = 0 and solve for y.
Thank you. Is there an official formula for what you came up with or some sort of rearranged formula?
To solve this problem, we can use the concept of heat transfer and the principle of conservation of energy. The heat gained by the ice cube as it melts is equal to the heat lost by the water in the cup (assuming no heat is lost to the surroundings). We can calculate the heat gained by the ice cube and the heat lost by the water using the specific heat capacities of ice and water, and then solve for the mass of the ice cube.
The formula for the heat transferred is:
Q = m * c * ΔT
where:
Q is the heat transferred in joules (J)
m is the mass in grams (g)
c is the specific heat capacity in joules per gram per degree Celsius (J/g°C)
ΔT is the change in temperature in degrees Celsius (°C)
The specific heat capacity of ice is approximately 2.09 J/g°C, and the specific heat capacity of water is approximately 4.18 J/g°C.
Let's consider the heat gained by the ice cube:
Q_ice = m_ice * c_ice * (T_final - T_initial)
where:
Q_ice is the heat gained by the ice cube
m_ice is the mass of the ice cube (what we want to calculate)
c_ice is the specific heat capacity of ice (2.09 J/g°C)
T_final is the final temperature of the mixture (20.0°C)
T_initial is the initial temperature of the ice cube (0°C)
Now, let's consider the heat lost by the water:
Q_water = m_water * c_water * (T_final - T_initial)
where:
Q_water is the heat lost by the water
m_water is the mass of the water (given as 30.0 g)
c_water is the specific heat capacity of water (4.18 J/g°C)
T_final is the final temperature of the mixture (20.0°C)
T_initial is the initial temperature of the water (45.0°C)
Since the ice cube melts, its final temperature will also be 0°C. The heat lost by the water is equal to the heat gained by the ice cube, so we can set up an equation:
Q_ice = Q_water
m_ice * c_ice * (T_final - T_initial) = m_water * c_water * (T_final - T_initial)
Now we can substitute the known values into the equation and solve for m_ice:
m_ice * 2.09 * (20 - 0) = 30.0 * 4.18 * (20 - 45)
Simplifying the equation:
m_ice * 41.8 = 30.0 * (-57.0)
Dividing both sides by 41.8:
m_ice = (30.0 * (-57.0)) / 41.8
m_ice ≈ -40.82 g
The mass of the ice cube is approximately -40.82 grams. However, mass cannot be negative, so it seems there might be an error in the calculations or the given information. Please double-check the values and equations used.