A sample of zinc with a mass of 14g was heated to 98 degree celsius in boiling water and then dropped into a beaker containing 4.5x10^4ng of water at 25 degree celsius. when the water and metal came to thermal equilibrium, the temperature is 29 degree celsius. what is the specific heat of zinc?
To solve this problem, we can use the principle of conservation of energy. The heat gained by the water equals the heat lost by the zinc. We can use the following equation:
Qwater = -Qzinc
To find Qwater, we need to calculate the heat gained by the water using the formula:
Qwater = mass of water * specific heat of water * change in temperature
Given:
Mass of water (mwater) = 4.5x10^4 ng = 4.5x10^-2 g (since 1 g = 10^9 ng)
Specific heat of water (cwater) = 4.184 J/g°C (approximately)
Change in temperature (ΔT) = (29 - 25) °C = 4 °C
Now we can substitute the values into the formula to find Qwater:
Qwater = (4.5x10^-2 g) * (4.184 J/g°C) * (4 °C)
Next, let's determine Qzinc. We can use the formula:
Qzinc = mass of zinc * specific heat of zinc * change in temperature
Given:
Mass of zinc (mzinc) = 14 g
Change in temperature (ΔT) = (98 - 29) °C = 69 °C
Now we can substitute the values into the formula to find Qzinc:
Qzinc = (14 g) * (czinc) * (69 °C)
Since we need to find the specific heat of zinc (czinc), we can rearrange the equation:
czinc = Qzinc / (mass of zinc * change in temperature)
Substituting the values, we get:
czinc = [(14 g) * (69 °C)] / Qzinc
Plugging the calculated values for Qwater and Qzinc, we can now solve for the specific heat of zinc (czinc).
To find the specific heat of zinc, we can use the principle of heat transfer. The heat lost by the zinc will be equal to the heat gained by the water.
The formula for heat transfer is:
Q = mcΔT
Where:
Q is the heat transferred (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity (in joules per kilogram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
Let's calculate the heat lost by the zinc:
We have the mass of the zinc (m1) = 14 g = 0.014 kg
The initial temperature of the zinc (T1) = 98°C
The final temperature of the zinc and water (T2) = 29°C
ΔT = T2 - T1 = 29°C - 98°C = -69°C
Now, let's calculate the heat gained by the water:
We have the mass of water (m2) = 4.5x10^4 ng = 0.045 kg
ΔT = T2 - T1 = 29°C - 25°C = 4°C
Since the heat lost by the zinc is equal to the heat gained by the water, we can set up the equation:
m1c1ΔT1 = m2c2ΔT2
Substituting the values we found:
0.014c1(-69) = 0.045c2(4)
Simplifying the equation:
-0.966c1 = 0.18c2
Now we can solve for the specific heat of zinc (c1):
c1 = (0.18c2) / -0.966
Since we want the answer in joules per kilogram per degree Celsius, let's convert the mass of water to kilograms:
m2 = 0.045 kg
Using the equation above, we can substitute the value of c2 and calculate c1. The specific heat of water is typically around 4186 J/kg°C.
c1 = (0.18 x 4186) / -0.966
Calculating the above expression, we find:
c1 ≈ -775.4 J/kg°C
The specific heat of zinc is approximately -775.4 J/kg°C.