A 43.0 g silver spoon at 22.3°C is placed in a cup of coffee at 90.0°C. The spoon absorbs heat from the coffee to reach a temperature of 89.0°C. If the silver spoon placed in the coffee causes it to cool 0.89°C, what is the mass of the coffee?
To solve this problem, we can use the principle of conservation of energy.
The heat absorbed by the spoon is equal to the heat lost by the coffee. The heat absorbed by the spoon can be calculated using the formula:
Q = mcΔT
Where:
Q = heat absorbed (or lost) in joules
m = mass of the object (spoon or coffee) in grams
c = specific heat capacity of the object in J/g°C
ΔT = change in temperature in °C
We know the following values:
For the spoon:
m_spoon = 43.0 g
c_silver = 0.237 J/g°C
ΔT_spoon = 89.0°C - 22.3°C = 66.7°C
For the coffee:
ΔT_coffee = 90.0°C - (89.0°C - 0.89°C) = 0.89°C
Let's first calculate the heat absorbed by the spoon:
Q_spoon = m_spoon * c_silver * ΔT_spoon
Q_spoon = 43.0 g * 0.237 J/g°C * 66.7°C
Q_spoon = 662.331 J
Since the heat absorbed by the spoon is equal to the heat lost by the coffee, we can set up the equation:
Q_spoon = Q_coffee
662.331 J = m_coffee * c_coffee * ΔT_coffee
We can rearrange this equation to solve for the mass of the coffee:
m_coffee = Q_spoon / (c_coffee * ΔT_coffee)
Let's assume the specific heat capacity for coffee is approximately 4.18 J/g°C (value for water).
m_coffee = 662.331 J / (4.18 J/g°C * 0.89°C)
m_coffee ≈ 177.783 g
Therefore, the mass of the coffee is approximately 177.783 grams.
To solve this problem, we can use the equation for heat transfer:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this case, the silver spoon absorbs heat from the coffee. Therefore, the heat transferred from the coffee to the spoon is equal to the heat absorbed by the spoon. We can set up the following equation:
(mcΔT)coffee = (mcΔT)spoon
The change in temperature, ΔT, can be calculated by subtracting the initial temperature from the final temperature:
ΔT = final temperature - initial temperature
Given:
Mass of the silver spoon (m) = 43.0 g
Initial temperature of the spoon (Tspoon initial) = 22.3°C
Final temperature of the spoon (Tspoon final) = 89.0°C
Change in temperature of the spoon (ΔTspoon) = Tspoon final - Tspoon initial = 89.0°C - 22.3°C = 66.7°C
Change in temperature of the coffee (ΔTcoffee) = 0.89°C (as given)
Now, we can rearrange the equation to solve for mass of the coffee (mcoffee):
mcoffee = [(mcΔT)spoon] / [(ΔTcoffee)]
Plugging in the values:
mcoffee = [(43.0 g × 66.7°C) / 0.89°C]
mcoffee = 43.0 g × 66.7 / 0.89
mcoffee ≈ 3221.91 g
Therefore, the mass of the coffee is approximately 3221.91 grams.