A student in a lab heats up 50 g of water using a Bunsen burner and an insulated cup to ensure no heat loss to surroundings. The water is initially at 25 °C and is heated until boiling at 100 °C. How much heat is transferred to the water in the insulated cup in the previous question? The specific heat capacity of water is 4.184 J/K/mol.
My answer was 1.7 but it's wrong. Can someone please help? Thanks
I don't know what the previous question is but for this one, you have
50 g x 4.184 J/g*K x (100-25) = 15,689 J which I would round to 1.6 kJ. Watch the significant figures
Note the Cp for H2O is 4.184 J/g and not J/mol
To calculate the amount of heat transferred to the water, you need to use the equation:
Q = m * c * ΔT
Where:
Q = Heat transferred (in joules)
m = Mass of the water (in grams)
c = Specific heat capacity of water (in J/g·°C)
ΔT = Change in temperature (in °C)
Given:
Mass of water (m) = 50 g
Specific heat capacity of water (c) = 4.184 J/g·°C
Initial temperature (T1) = 25 °C
Final temperature (T2) = 100 °C
Now, calculate the change in temperature:
ΔT = T2 - T1
ΔT = 100 °C - 25 °C
ΔT = 75 °C
Next, substitute the values into the equation to calculate the heat transferred:
Q = m * c * ΔT
Q = 50 g * 4.184 J/g·°C * 75 °C
Calculating further:
Q = 2092 J
Therefore, the correct answer is that 2092 joules of heat were transferred to the water in the insulated cup.
To calculate the amount of heat transferred to the water, you can use the formula:
Q = m × c × ΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
In this case, the mass of the water (m) is 50 g, the specific heat capacity of water (c) is 4.184 J/K/mol, and the change in temperature (ΔT) is the difference between the final and initial temperatures, which is 100 °C - 25 °C = 75 °C.
However, before using this formula, we need to convert grams to moles because the specific heat capacity of water is given in J/K/mol.
To convert grams to moles, you can use the molar mass of water, which is approximately 18 g/mol.
mol = mass (g) ÷ molar mass (g/mol)
= 50 g ÷ 18 g/mol
≈ 2.78 mol
Now we can calculate the heat transferred:
Q = m × c × ΔT
= 2.78 mol × 4.184 J/K/mol × 75 °C
≈ 876 J (rounded to three significant figures)
So the amount of heat transferred to the water in the insulated cup is approximately 876 J.