A 140-W heater was placed in 2.00 kg of methanol and turned on for exactly 1 min. The temperature increased by 3.54°C. Assuming that all the heat is absorbed by the methanol, calculate the molar heat capacity of methanol.
To calculate the molar heat capacity of methanol, we can use the formula:
q = mcΔT
Where:
q = heat absorbed by the methanol (in Joules)
m = mass of the methanol (in kg)
c = molar heat capacity of methanol (in J/mol°C)
ΔT = change in temperature (in °C)
We are given:
q = 140 W (power of the heater) * 1 min (converted to seconds) = 140 J/s * 60 s = 8400 J
m = 2.00 kg
ΔT = 3.54°C
Plugging in these values into the formula, we have:
8400 J = (2.00 kg) * c * (3.54°C)
To solve for c, we rearrange the equation:
c = 8400 J / [(2.00 kg) * (3.54°C)]
c = 1191.49 J / (kg°C)
Therefore, the molar heat capacity of methanol is approximately 1191.49 J/(mol°C).
To calculate the molar heat capacity of methanol, we can use the equation:
q = mcΔT
Where:
q is the heat absorbed or released
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's determine the heat absorbed by methanol (q):
q = P × t
Where:
P is the power of the heater (given as 140 W)
t is the time in seconds (given as 1 min = 60 s)
q = 140 W × 60 s
q = 8400 J
Now, let's substitute the values into the equation for heat:
8400 J = (2.00 kg) × c × (3.54 °C)
To find the molar heat capacity, we need to convert the mass of methanol from kilograms to moles. We can use the molar mass of methanol (CH3OH), which is 32.04 g/mol.
2.00 kg = 2000 g
2000 g ÷ 32.04 g/mol = 62.45 mol
Now, we can solve for c:
8400 J = (62.45 mol) × c × (3.54 °C)
To isolate c, divide both sides of the equation by (62.45 mol × 3.54 °C):
c = 8400 J / (62.45 mol × 3.54 °C)
Now, we can calculate the molar heat capacity of methanol by evaluating the expression above:
c ≈ 39.33 J/(mol·°C)
Therefore, the molar heat capacity of methanol is approximately 39.33 J/(mol·°C).
Heat produced: 140*60 joules
heatproduced=2kg*c*3.54
calculate c.