Final temperature of the liquid water?

[name_ask17]

Homework Statement

A 10-kg block of ice has a temperature of -12°C. The pressure is one atmosphere. The block absorbs 4.15 106J of heat. What is the final temperature of the liquid water?

Homework Equations

Q=mcT

The Attempt at a Solution

i know i have to find the change from -12 to 0 deg c and then use mLfusion for the ice and add them. i am lost after that. i keep getting the wrong answer. how do i do this?

 

[rock.freak667]

The Attempt at a Solution

i know i have to find the change from -12 to 0 deg c and then use mLfusion for the ice and add them. i am lost after that. i keep getting the wrong answer. how do i do this?

You'll need to do an energy balance.

Heat absorbed = heat to change phase of the ice + heat to increase the temperature from -12 to T.

you know that heat to change phase of the ice=mL_fusion and the heat to increase the temperature = mc(T₂-T₁).

 

[name_ask17]

so would i do Q=mcT+mL as 4.15E6=(10)(4186)(T-12)+(10)(334)? because that is not giving me the correct answer. can you please explain to me what I am doing wrong here?

 

[name_ask17]

dont bother. i undersand it now. i was to take the heat gained to ive, ice melting, and water warming up. thank you for your assistance

 

[rwisz]

I would approach this problem by first identifying the known values and the unknown value. In this case, we know the mass of the ice (10 kg), its initial temperature (-12°C), and the amount of heat absorbed (4.15 x 10^6 J). The unknown value is the final temperature of the liquid water.

Next, I would use the equation Q=mcT to calculate the change in temperature of the ice. We know the specific heat capacity of ice (c) is 2.1 J/g°C and the mass (m) is 10 kg, so we can plug in these values to solve for the change in temperature (T).

Q = (10 kg)(2.1 J/g°C)(T - (-12°C))

4.15 x 10^6 J = (10 kg)(2.1 J/g°C)(T + 12°C)

T + 12°C = 1976190.48

T = 1976190.48 - 12°C

T = 1976178.48°C

This is the change in temperature of the ice. To find the final temperature of the liquid water, we need to add this change to the initial temperature of the ice.

Final temperature = -12°C + 1976178.48°C = 1976166.48°C

However, this answer is not realistic as it is well above the boiling point of water. This indicates that the ice has not completely melted and there may be some error in the calculations.

To find the final temperature more accurately, we need to take into account the latent heat of fusion for ice, which is the energy required to melt the ice without changing its temperature. This can be calculated using the equation Q=mLfusion, where m is the mass of the ice and Lfusion is the latent heat of fusion for ice (334 kJ/kg).

Q = (10 kg)(334 kJ/kg)

Q = 3340 kJ

Now, we can subtract this amount of energy from the total heat absorbed to find the remaining energy that will be used to increase the temperature of the liquid water.

Remaining energy = 4.15 x 10^6 J - 3340 kJ = 4.14666 x 10^6 J

Using the equation Q=mcT, we can now solve for the final temperature of the liquid water.

4.14666 x 10