How Much Energy Is Needed to Freeze Water and Melt Ice Through Friction?

[tbwtki]

Help Please!

Hello Could someone help me with this problem? thanks

Part
A:
A container holds 35.0 kg of water at 20.0 C. HOw much energy must be removed from the water to turn it completely into ice at 0 C?

Part
B:
The block of ice is projected a cross a surface at an initial speed of 6.50m/s. While it is sliding across the surface, the friction force casues the ice to melt. After the block has stopped sliding and come to rest, how much ice has melted? Assume that all tof the internal enrgy transformed by the frction force goes into the ice -- none goes into the surface.

C:
With the same assumptions as in part (b), with what speed would the block of ice have to be projected inorder to completely melt it due to friction?

 

[Chi Meson]

I recognize this question. It's from an AP "B" exam from the early 90's.

First off, remember that heat is energy. One formula (Q = c m deltaT) will tell you how much heat must be removed to get the water to zero degrees. A second formula (Q = mL) will tell you how much heat must be removed to turn that zero degree liquid water into zero degree ice water.

Second part: kinetic energy of ice is assumed to turn into heat. HOw much KE? use that amount in that second formula above to see how much mass of ice melts.

Third. HOw much total KE is needed to melt all the ice? the same as the energy needed to turn the water into ice (second half of the first part). What speed corresponds to that amount of KE?

 

[M98Ranger]

Hello! I can definitely assist you with your energy physics homework. Let's take a look at each part separately.

Part A: To turn water into ice, we need to remove energy from it. The amount of energy required can be calculated using the equation Q = m * L, where Q is the energy, m is the mass, and L is the specific latent heat of fusion (for water, this is 334 kJ/kg). So, in this case, we have Q = 35.0 kg * 334 kJ/kg = 11,690 kJ. Therefore, 11,690 kJ of energy must be removed from the water to turn it into ice at 0 C.

Part B: In this part, we need to determine how much ice has melted after the block has come to rest. To do this, we first need to calculate the kinetic energy of the block using the equation KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the speed. So, in this case, we have KE = 1/2 * 35.0 kg * (6.50 m/s)^2 = 1,126.25 J. This is the amount of energy that goes into melting the ice. Now, we can use the equation Q = m * L to find the mass of ice that has melted. So, Q = m * 334 kJ/kg, and we know Q = 1,126.25 J. Solving for m, we get m = 1,126.25 J / 334 kJ/kg = 0.0034 kg. Therefore, 0.0034 kg of ice has melted due to friction.

Part C: Finally, we need to find the initial speed at which the block of ice would have to be projected to completely melt it due to friction. Using the same equation as before, Q = m * L, we can solve for v by plugging in the values we know. So, 11,690 kJ = m * 334 kJ/kg, and we know m = 35.0 kg. Solving for v, we get v = √(11,690 kJ / 35.0 kg * 334 kJ/kg) = 11.6 m/s. Therefore, the block of ice would need to be projected at an initial