At a local convenience store, you purchase a cup of coffee, but, at 98.4oC, it is too hot to drink. You add 23.0 g of ice that is –2.2oC to 248 mL of the coffee. What is the final temperature of the coffee? (Assume the heat capacity and density of the coffee are the same as water and the coffee cup is well insulated.)
To determine the final temperature of the coffee after adding ice, we can use the principles of heat transfer and conservation of energy. Here's how you can calculate it step by step:
Step 1: Calculate the heat lost by the hot coffee.
The heat lost by the hot coffee can be calculated using the formula:
Q1 = m1 * c * ΔT1
Where:
Q1 is the heat lost by the hot coffee (in joules)
m1 is the mass of the coffee (in grams)
c is the specific heat capacity of water (4.184 J/g°C)
ΔT1 is the change in temperature of the coffee (final temperature - initial temperature)
Given:
m1 = 248 mL = 248 g (since the density of water and coffee are assumed to be the same)
ΔT1 = Final Temperature - Initial Temperature = Unknown
Step 2: Calculate the heat gained by the ice.
The heat gained by the ice can be calculated using the formula:
Q2 = m2 * c * ΔT2
Where:
Q2 is the heat gained by the ice (in joules)
m2 is the mass of the ice (in grams)
c is the specific heat capacity of water (4.184 J/g°C)
ΔT2 is the change in temperature of the ice (final temperature - initial temperature)
Given:
m2 = 23.0 g
ΔT2 = Final Temperature - Initial Temperature (Unknown)
Step 3: Equate the heat lost and the heat gained.
Since the heat lost by the coffee is equal to the heat gained by the ice (assuming no heat is lost to the surroundings), we can set up the equation:
Q1 = Q2
m1 * c * ΔT1 = m2 * c * ΔT2
Substituting the known values:
248 * 4.184 * (Final Temperature - 98.4) = 23.0 * 4.184 * (Final Temperature - (-2.2))
Simplifying the equation, you will be able to solve for the Final Temperature.
To determine the final temperature of the coffee after adding ice, we can use the principle of conservation of energy.
1. First, let's calculate the heat gained by the ice:
Q_ice = m_ice * C_ice * ΔT_ice
where:
- m_ice is the mass of the ice = 23.0 g
- C_ice is the specific heat capacity of ice = 2.09 J/g·°C
- ΔT_ice is the change in temperature of the ice = final temperature of the ice - initial temperature of the ice
Since the ice initially is at -2.2°C and we want to determine the final temperature, ΔT_ice = Tf_ice - (-2.2).
Since we are assuming the ice melts completely, Tf_ice will be 0°C.
Therefore, ΔT_ice = 0 - (-2.2).
Substituting the values:
Q_ice = 23.0 g * 2.09 J/g·°C * (-2.2 - (-2.2))
Q_ice = 0 J (since the change in temperature is zero)
2. Next, let's calculate the heat lost by the coffee:
Q_coffee = m_coffee * C_water * ΔT_coffee
where:
- m_coffee is the mass of the coffee = 248 mL (or 248 g, since the density of water is 1 g/mL)
- C_water is the specific heat capacity of water = 4.18 J/g·°C (approximately)
- ΔT_coffee is the change in temperature of the coffee = final temperature of the coffee - initial temperature of the coffee
We can rewrite the equation as:
Q_coffee = m_coffee * C_water * ΔT_coffee
Q_coffee = 248 g * 4.18 J/g·°C * (Tf_coffee - 98.4)
3. Since the system is insulated, the heat lost by the coffee is equal to the heat gained by the ice. So we can equate the two equations:
Q_ice = Q_coffee
0 J = 248 g * 4.18 J/g·°C * (Tf_coffee - 98.4)
4. Simplify and solve for Tf_coffee:
0 = 1035.84 g·°C * (Tf_coffee - 98.4)
0 = Tf_coffee - 98.4
Tf_coffee - 98.4 = 0
Tf_coffee = 98.4°C
Therefore, the final temperature of the coffee after adding the ice is 98.4°C.