In a calorimeter, 100 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g. How much heat was absorbed?
33.4 J
334 J
33.4 kJ
334 kJ
.334kJ/g
.334 * 100 = 33.4 kJ
How much heat (in kilojoules) is evolved or absorbed in the reaction of 1.10g of Na with H 2 O ?
2Na(s)+2H 2 O(l)�¨2NaOH(aq)+H 2 (g) ƒ¢H ∘ = -368.4kJ .
To calculate the amount of heat absorbed, you can use the formula:
Q = mcΔT
Where:
Q = amount of heat absorbed
m = mass of the substance (in grams)
c = specific heat capacity of the substance
ΔT = change in temperature
Since the ice is melting at 0°C, the change in temperature is 0°C. The specific heat capacity for the phase change of ice is also known as the enthalpy of fusion, which is given as 334 J/g.
Plugging in the values into the formula:
Q = (100g) x (334 J/g) x (0°C - 0°C)
Q = 0 J
Therefore, the amount of heat absorbed is 0 J.
To find the amount of heat absorbed, we can use the formula:
Q = m * ΔHf
Where:
Q is the amount of heat absorbed
m is the mass of the substance (ice in this case)
ΔHf is the enthalpy of fusion of the substance (334 J/g for ice)
Given that the mass of the ice is 100 g and the enthalpy of fusion of ice is 334 J/g, we can calculate the amount of heat absorbed as follows:
Q = 100 g * 334 J/g
Q = 33400 J
Therefore, the amount of heat absorbed is 33.4 kJ (converted from J).