DrBob222,HELP!
Two 20.0-g ice cubes at –10.0 °C are placed into 275 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.
heat capacity h2o (S)=37.7 j/mol*k
heat capacity h2o (l)=75.3 j/mol*k
enthalpy of fusion of h2o= 6.01 kj/mol
could you elaberate more with this. im still really cofused! thanks!
Substitute those numbers into the equation I gave you and solve for Tfinal. That is the only unknown in the equation; you will have numbers for everything else.
For example the first part was
mass ice x specific heat ice x (Tfinal-Tinitiaol).
mass ice = 40 g. Convert to mol(that'sw necessary because you have listed specific heat in J/mol*C). mol = grams/molar mass = 40/18 = 2.22 mol.
specific heat ice = 37.7 J/mol
Tfinal = unknown
Tinitial = -10
(2.22 x 37.7 x [Tf-(-10)] then do the next part etc.
It's nothing but arithmetic from here on out.
Sure, I'd be happy to help explain the steps involved in solving this problem.
To find the final temperature of the water after all the ice melts, you need to consider the heat transfer between the ice and the water.
Here's how to approach the problem:
1. Calculate the heat absorbed or released by each component using the formula Q = m * C * ΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
For the ice cubes:
Q_ice = m_ice * C_ice * ΔT_ice
For the water:
Q_water = m_water * C_water * ΔT_water
2. Calculate the heat absorbed or released during the phase change of the ice. This is called the heat of fusion and is given by the formula Q_fusion = m_ice * ΔH_fusion.
3. Since the system is isolated (no energy is transferred to or from the surroundings), the heat lost by the water when the ice cools down is equal to the heat gained by the ice to melt. So, we can set up the equation:
Q_water + Q_fusion = Q_ice
4. Rearrange the equation to solve for the final temperature of the water (ΔT_water):
ΔT_water = (Q_ice - Q_fusion) / (m_water * C_water)
5. Plug in the given values into the appropriate equations:
m_ice = 20.0 g
ΔT_ice = 0 °C - (-10.0 °C) = 10.0 °C
m_water = 275 g
ΔT_water = ?
C_ice = 75.3 J/mol*K (since the enthalpy of fusion is given, it is for 1 mole of water, and the heat capacity is per mole as well)
C_water = 4.18 J/g*K (specific heat capacity of water)
ΔH_fusion = 6.01 kJ/mol = 6.01 × 10^3 J / (18.01528 g/mol) (conversion factor from kJ to J and gram to mole)
6. Substitute the given values into the equations and solve for the final temperature of the water.
I hope this explanation helps! If you have any more specific questions, feel free to ask.