A 20 g sample of aluminum at 85 degree is dropped in 500g of water. The resulting temperature of aluminum and water is 25 degree. What is the initial temperature of water?
Set the heat lost by aluminum equal to the heat gained by water, and solve for the initial water temperture To (the only unknown). You will need the specific heat of aluminum, C(al).
20 (85 - 25) *C(al) = (25-To)*C(water)*500
To find the initial temperature of water, we can use the principle of conservation of heat energy.
The heat gained by the water is equal to the heat lost by the aluminum.
The equation used to calculate the heat absorbed or released in a system is given by:
Heat gained or lost = mass * specific heat capacity * change in temperature.
First, let's calculate the heat lost by the aluminum:
Heat lost by aluminum = (mass of aluminum) * (specific heat capacity of aluminum) * (change in temperature of aluminum)
Given:
mass of aluminum (m_aluminum) = 20g
specific heat capacity of aluminum (c_aluminum) = 0.897 J/g°C (source: https://www.engineeringtoolbox.com/specific-heat-metals-d_152.html)
change in temperature of aluminum (ΔT_aluminum) = (final temperature of aluminum - initial temperature of aluminum) = (25°C - 85°C) = -60°C (as the temperature has decreased)
Substituting the values into the equation, we have:
Heat lost by aluminum = (20g) * (0.897 J/g°C) * (-60°C)
Next, let's calculate the heat gained by the water:
Heat gained by water = (mass of water) * (specific heat capacity of water) * (change in temperature of water)
Given:
mass of water (m_water) = 500g
specific heat capacity of water (c_water) = 4.18 J/g°C (source: https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html)
change in temperature of water (ΔT_water) = (final temperature of water - initial temperature of water)
We need to solve for the initial temperature of water. Let's call it T_water_initial.
Heat gained by water = (500g) * (4.18 J/g°C) * (T_water_final - T_water_initial)
Since the heat gained by the water is equal to the heat lost by the aluminum (assuming no heat loss to the surrounding):
Heat gained by water = Heat lost by aluminum
Therefore, we have the equation:
(500g) * (4.18 J/g°C) * (T_water_final - T_water_initial) = (20g) * (0.897 J/g°C) * (-60°C)
Now we can solve for T_water_initial:
(500g) * (4.18 J/g°C) * (T_water_final - T_water_initial) = (20g) * (0.897 J/g°C) * (-60°C)
Divide both sides of the equation by (500g) * (4.18 J/g°C):
T_water_final - T_water_initial = (20g) * (0.897 J/g°C) * (-60°C) / (500g) * (4.18 J/g°C)
T_water_final - T_water_initial = -2.1652
Finally, rearrange the equation and solve for T_water_initial:
T_water_initial = T_water_final - (-2.1652)
T_water_initial = T_water_final + 2.1652
Now we can substitute the given final temperature of the system to find the initial temperature of water.