A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 79°C. The other sample has a mass of 83 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample.
I know the equation is Q = Q for this but when i set it up and put the answer in, it says i am 10% from the right answer.
In my equation i ignored the mass and specific of thealuminum calorimeter.
so am I supposed to include that into my equation?
if so..how?
yes, add the calorimeter. It has a mass of 100 g, specific heat of aluminum intitial temp of 10C, final temp is 20C.
To determine the specific heat of the unknown second sample, you need to consider the heat gained or lost by each component in the system and equate them.
Let's divide the problem into three parts: the water, the copper block, and the unknown second sample.
1. The heat gained or lost by the water can be calculated using the equation:
Q_water = mass_water * specific_heat_water * change_in_temperature_water
Given:
mass_water = 250 g
specific_heat_water = 4.18 J/g°C (specific heat of water)
change_in_temperature_water = final temperature - initial temperature = 20°C - 10°C = 10°C
So, Q_water = 250 g * 4.18 J/g°C * 10°C = 10,450 J
2. The heat gained or lost by the copper block can be calculated using the equation:
Q_copper = mass_copper * specific_heat_copper * change_in_temperature_copper
Given:
mass_copper = 50 g
specific_heat_copper = 0.385 J/g°C (specific heat of copper)
change_in_temperature_copper = final temperature - initial temperature = 20°C - 79°C = -59°C (negative because it loses heat)
So, Q_copper = 50 g * 0.385 J/g°C * (-59°C) = -1,120.15 J (note the negative sign indicates heat loss)
3. Now let's consider the unknown second sample. Let's assume its specific heat is given by the symbol "specific_heat_unknown".
The heat gained or lost by the unknown second sample can be calculated using the equation:
Q_unknown = mass_unknown * specific_heat_unknown * change_in_temperature_unknown
Given:
mass_unknown = 83 g
change_in_temperature_unknown = final temperature - initial temperature = 20°C - 100°C = -80°C (negative because it loses heat)
So, Q_unknown = 83 g * specific_heat_unknown * (-80°C)
To find the specific heat of the unknown second sample, we can equate the total heat gained by the water, copper block, and unknown second sample. So:
Q_water + Q_copper + Q_unknown = 0 (Since the system stabilizes, 0 heat is gained or lost by the entire system)
Substituting the values we calculated for Q_water and Q_copper, we get:
10,450 J - 1,120.15 J + 83 g * specific_heat_unknown * (-80°C) = 0
Now, you can solve this equation for the specific heat of the unknown second sample (specific_heat_unknown). Once you calculate it, plug in the value and check if it matches the correct answer.