what is the mass of a lead block that if heated to 100.0 dgrs C and placed in a container with 50.0g of ice at 0.00dgrs C produces a final water sample of 15.0dgrs C? Sh of Pb= 0.128J/g dgrsC step by step solution and answer
q1 for Pb block = mass x sp.h. x (Tf - Ti).
mass Pb = X
sp.h. Pb = listed
Tf = 15.0
Ti = 100.0
q2 for ice at zero degrees to melt to water at zero degrees is mass x delta H fusion. Be sure to use J/g for the unit.
mass = 50.0 g
delta H = ?? J/g.
q3 for water at zero to be heated to 15.0 degrees = mass x sp.h. x (Tf - Ti)
mass H2O = 50 g.
sp.h. water = ?? J/g*C.
Tf = 15.0
Ti = 0.0
q1+q2+q3 = 0
solve for X.
The sum of the heats gained is zero (some lose heat).
Heat gained lead+ heatgainedice=0
m*Cpb*(100-Tf)+ 50*Heatfusionice + 50*cwater*(Tf-0)=0
you are given Tf, solve for m.
To find the mass of the lead block, we can use the equation:
q1 = mass x specific heat x (Tf - Ti)
Here, q1 represents the heat gained by the lead block, mass is the mass of the lead block (which we are trying to find), specific heat is the specific heat capacity of lead (0.128 J/g°C), Tf is the final temperature (15.0°C) and Ti is the initial temperature (100.0°C).
Next, let's find q2, which represents the heat gained by the ice to melt it into water. We can use the equation:
q2 = mass x ΔHfusion
Where mass is the mass of the ice (50.0 g) and ΔHfusion is the heat of fusion for ice (the energy required to convert 1 gram of ice into water at 0.00°C), which we need to determine.
Lastly, let's find q3, which represents the heat gained by the water to raise its temperature from 0.00°C to 15.0°C. We can use the equation:
q3 = mass x specific heat x (Tf - Ti)
Here, mass is the mass of the water (50 g), specific heat is the specific heat capacity of water (which we need to determine), Tf is the final temperature (15.0°C), and Ti is the initial temperature (0.00°C).
Now, using the principle that the sum of the heats gained is zero, we can set up the equation:
q1 + q2 + q3 = 0
Substituting the corresponding expressions, we get:
mass x specific heat x (100.0 - 15.0) + 50.0 x ΔHfusion + 50.0 x specific heat x (15.0 - 0.0) = 0
We are given the values for Tf and Ti, and we need to solve for the mass of the lead block (X).
By rearranging the equation and solving for X, we can determine the mass of the lead block.