A styrofoam cup with a mass of 2.97g has 81.68g of hot water at60.0 •C has ice added to it untill it reaches a temrature of 7.0•C and weights 141.12g. What is the molar heat lost by the water to melt the ice.
Equation: -(mcdeltaT)=(deltaHn)+(mcdeltaT)
m1=81.68g. DeltaH=? m=141.12g
c=4.19J/g•C. n=18.0152mol. c=4.19
DeltaT=-53•C DeltaT=53•C
I don't understand the trouble here. You have the equation and the numbers; all you do is substitute the numbers and turn the crank. Out comes the answer.
nadjima ha hol
To find the molar heat lost by the water to melt the ice, we can use the equation you provided: -(m1cΔT) = ΔHn + (m2cΔT), where:
m1 is the mass of water before adding ice,
c is the specific heat capacity of water,
ΔT is the change in temperature,
ΔH is the enthalpy change of the ice, and
n is the number of moles of ice.
Given values:
m1 = 81.68 g
c = 4.19 J/g·°C
ΔT = -53 °C (negative because the temperature decreased from 60.0°C to 7.0°C)
m2 = 141.12 g
First, let's calculate the heat lost by the water:
-(m1cΔT) = -(81.68 g)(4.19 J/g·°C)(-53 °C) = 18,545.5 J
Now, let's calculate the heat gained by the ice:
(m2cΔT) = (141.12 g)(4.19 J/g·°C)(-53 °C) = -31,250.07 J
Next, we can rearrange the equation to solve for ΔH:
ΔH = -(m1cΔT) - (m2cΔT) / n
ΔH = (18,545.5 J) - (-31,250.07 J) / 18.0152 mol
ΔH = 13,704.57 J / 18.0152 mol
ΔH = 760.68 J/mol (rounded to four significant figures)
Therefore, the molar heat lost by the water to melt the ice is approximately 760.7 J/mol.