A 55.0 g ice cube, initially at 0°C, is dropped into a Styrofoam cup containing 318 g of water, initially at 18.2°C. What is the final temperature of the water, if no heat is transferred to the Styrofoam or the surroundings?
Heat of fusion is the amount of heat energy required to change the state of a substance from solid to liquid.
λ = 335000 J/kg is heat of fusion of water-ice
c = 4185.5 J/(kg•K) is heat capacity of water
Q1 = λ•m1
Q2 =c•m1 •(t-t1) =c•m1•t
Q3 = c•m2 •(t2-t) .
Q1+Q2 = Q3
λ•m1 + c•m1•t = c•m2 •(t2-t),
λ•m1 + c•m1•t = c•m2 •t2 - c•m2 •t,
c•m1•t+ c•m2 •t = c•m2 •t2 - λ•m1
t = (c•m2•t2 - λ•m1)/[c•(m1+m2)] =( 4 180•0.318•18.2 -335000•0.055)/[4180•(0.335•0.055)] = 3.7 oC
To find the final temperature of the water, we can use the concept of heat transfer. The heat gained by the water will be equal to the heat lost by the ice cube.
The heat gained by the water can be calculated using the formula:
Q_water = m_water * C_water * ΔT
Where:
- Q_water is the heat gained by the water
- m_water is the mass of the water
- C_water is the specific heat capacity of water
- ΔT is the change in temperature of the water
The heat lost by the ice can be calculated using the formula:
Q_ice = m_ice * C_ice * ΔT
Where:
- Q_ice is the heat lost by the ice
- m_ice is the mass of the ice
- C_ice is the specific heat capacity of ice
- ΔT is the change in temperature of the ice
Since no heat is transferred to the surroundings or the Styrofoam cup, the heat gained by the water will be equal to the heat lost by the ice:
Q_water = Q_ice
m_water * C_water * ΔT_water = m_ice * C_ice * ΔT_ice
Since the ice is initially at 0°C, ΔT_ice = 0 - T_final (where T_final is the final temperature of the water)
Substituting the given values:
(318 g) * (4.18 J/g°C) * (T_final - 18.2°C) = (55.0 g) * (2.09 J/g°C) * (0°C - T_final)
Simplifying the equation:
1331.08 * T_final - 10438.76 = -115.00 * T_final
146.00 * T_final = 10438.76
T_final = 71.5°C (rounded to one decimal place)
Therefore, the final temperature of the water is 71.5°C.
To find the final temperature of the water, we can use the principle of conservation of energy.
First, let's calculate the heat gained by the ice cube when it melts. The heat gained by the ice can be calculated using the equation:
q = m * L_f
where q is the heat gained, m is the mass of the ice cube (55.0 g), and L_f is the heat of fusion for water (334 J/g).
q = (55.0 g) * (334 J/g)
q = 18370 J
This is the amount of heat that the ice cube absorbs from the water.
Next, we need to calculate the amount of heat lost by the water. To do this, we can use the specific heat capacity equation:
q = m * c * ΔT
where q is the heat lost, m is the mass of the water (318 g), c is the specific heat capacity of water (4.18 J/(g·°C)), and ΔT is the change in temperature.
Since the heat lost by the water is equal to the heat gained by the ice cube (assuming no heat is transferred to the surroundings), we can set up the equation:
(18370 J) = (318 g) * (4.18 J/(g·°C)) * (final temperature - 18.2°C)
Now, let's solve for the final temperature:
(18370 J) / (318 g * 4.18 J/(g·°C)) = final temperature - 18.2°C
57.8°C = final temperature - 18.2°C
final temperature = 57.8°C + 18.2°C
final temperature ≈ 76.0°C
Therefore, the final temperature of the water will be approximately 76.0°C if no heat is transferred to the Styrofoam or the surroundings.