A 50.0g sample of copper at 10.0 C is placed in 200.0g H20 at 40.0 C in a calorimeter. Heat flows from the water to the copper until both are at 39.3 C. What is the specific heat of copper? (Csp H20 is 4.18J/g*C.)
I don't have any idea how to do this problem. If someone could help me, it would be greatly appreciated. Thank you:)
heat to warm copper=
Ccopper * 50 *(39.3-10)
heat to cool water=
4.18 * 200 * (40-39.3)
set them equal and solve for Ccopper
To solve this problem, you need to use the concept of heat transfer and the equation:
q = m * c * ΔT
Where:
q is the heat transferred (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·°C),
ΔT is the change in temperature (in °C).
Let's calculate it step by step:
Step 1: Calculate the heat gained by the water.
q1 = m1 * c1 * ΔT1
m1 = mass of water = 200.0 g
c1 = specific heat capacity of water = 4.18 J/g·°C
ΔT1 = change in temperature = (39.3 °C - 40.0 °C) = -0.7 °C (as heat flows from hot to cold)
q1 = 200.0 g * 4.18 J/g·°C * -0.7 °C = -586.0 J
Note: The negative sign indicates the heat loss from the water.
Step 2: Calculate the heat gained by copper.
q2 = m2 * c2 * ΔT2
m2 = mass of copper = 50.0 g
c2 = specific heat capacity of copper (to be determined)
ΔT2 = change in temperature = (39.3 °C - 10.0 °C) = 29.3 °C
q2 = 50.0 g * c2 * 29.3 °C
Step 3: Set up the equation using the principle of energy conservation.
Since energy is conserved, the heat lost by the water must be equal to the heat gained by the copper.
q1 = q2
-586.0 J = 50.0 g * c2 * 29.3 °C
Step 4: Solve for the specific heat capacity of copper.
c2 = -586.0 J / (50.0 g * 29.3 °C)
c2 ≈ -0.400 J/g·°C
Note: The negative sign indicates a different direction of heat transfer compared to the water.
However, it doesn't make physical sense to have a negative specific heat capacity. Therefore, we assume that there was an error or negative sign in the calculations.
By correcting the values and considering the heat gained by the copper, we find:
c2 = 586.0 J / (50.0 g * 29.3 °C)
c2 ≈ 0.400 J/g·°C
Therefore, the specific heat capacity of copper is approximately 0.400 J/g·°C.
To solve this problem, we can use the principle of heat transfer, which states that heat lost by one substance is equal to the heat gained by another substance.
In this case, the heat lost by the water is equal to the heat gained by the copper. The formula we can use is:
qwater = -qCopper
where q represents the heat transfer and the negative sign indicates the direction of heat flow.
To calculate the heat transfer for each substance, we can use the equation:
q = m * C * ΔT
where q is the heat transfer, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
For water:
qwater = mwater * Csp H20 * ΔTwater
qwater = (200.0g) * (4.18J/g°C) * (39.3°C - 40.0°C)
For copper:
qCopper = mCopper * Csp Copper * ΔTCopper
qCopper = (50.0g) * Csp Copper * (39.3°C - 10.0°C)
Since the heat lost by the water is equal to the heat gained by the copper, we can set them equal to each other:
qwater = -qCopper
(200.0g) * (4.18J/g°C) * (39.3°C - 40.0°C) = -(50.0g) * Csp Copper * (39.3°C - 10.0°C)
Now we can solve for Csp Copper:
Csp Copper = [(200.0g) * (4.18J/g°C) * (39.3°C - 40.0°C)] / [(50.0g) * (39.3°C - 10.0°C)]
Plugging in the values and calculating, we find:
Csp Copper ≈ 0.39 J/g°C
Therefore, the specific heat of copper is approximately 0.39 J/g°C.