can you help me with this problem and show me step by step how to do it?

The changes that are happening.

1. Those requiring heat:

a) ice melts.

The heat spent by that is Q1 = m*L, where m is the mass of ice, and L is the heat of fusion.

b) the water due to that ice continually heats up from 0° until it reaches 15° and all ice has been melted.

This spent heat equals Q2 = m*c * Δt1. The m is still the mass of the ice (now turned to water), and t1 is the increase in temperature of that part of water (+ 15°C). c is the specific heat capacity of water.

2) Those giving off heat

The only process that gives off heat is the cooling of the water that was in the bottle before the ice was dropped

Q3 = m(2) * c * Δt(2)

m(2) is the mass of water initially in glass. c is again the specific heat capacity of water. Δt(2) = - 40 °C.

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Now that you got all that set, the idea is that when you add up all those heats, they will equal 0 (because what ice and heating water spends, the cooling water provides exactly)

Q1 + Q2 + Q3 = 0

Now it's just working out.

m*L + m*c * Δt1 + m(2) * c * Δt(2) = 0

You know m is the unknown. Do what you do.

m = - [ m(2) * c * Δt(2) ] / [ L + c * Δt1 ]

The last thing you should do before plugging in is convert everything equivalent units, so turn 6.01 kJ/mol into J/(gC) or something.

Should be m = 27.1 g if I didn't miss a digit somewhere - double-check the working :)

Best wishes!

There are a couple steps here. First, you have the given of how many jewels of energy are required to raise the temperature of water (or lower it) by a single degree celsius. So, since we know the temperature change of the water was 55 to 15, you can calculate the total energy that left the water and transferred to the ice.

You now have the energy that transferred to the ice.

What you now need to do is see how much ice can be both melted AND raised to 15 degrees using the energy available. Note, you will need to do some converting between grams and moles on both sides of the math. I recommend the goal post method, as my chemistry teacher called it. When you write your equations, write them as fractions and cross out negating numerator and denominator symbols until your answer is in the form you want. First see how much energy is needed for the single gram. Then see how many grams you can do this to using the energy you have.

Leave a comment if you need further explanation somewhere.

Conservation of energy....

"Step by step" solution to a heat exchange problem.

heat lost by water = heat gained by melting ice + heat gained by "ice water"

"ice water" is the water from the melted ice.

q = mcΔT ..... for changes in temperature

q = mHf ........ for phase change (fusion, aka melting)

The initial temperature of ice water is 0C. Both the hot water and the ice water have the same Tf, which is 15C.

The heat of fusion of ice is 334 J/g. (*)

-q(lost by hot water) = q(gained by ice and ice water)

-mcΔT(hot water) = mHf + mcΔT(ice water)

-mc(Tf - Thw) = mHf(ice) + mc(Tf - Tiw)

-64.3g x 4.18J/bC x 40C = m(334 J/g + 4.18J/gC x -15C)

m = 27.1g

=========

1 mol H2O x (6010 J / 1 mol H2O) x (1 mol H2O / 18.0g H2O) = 334 J/gC

========= Follow up ========

In Eveanne's lengthy solution, she omitted the heat needed to take the water from the melted ice to the final temperature of 15C. Therefore, it is of little use for you.

According with the Law of Conservation of matter,

Heat gained by Ice = +q = Heat lost by water = –q

or

+q = –q

Since

+q = + [(mass of ice)(heat of fusion of ice)]

and

–q = – [(mass of water)(specific heat of water)(change in temperature of water)]

then

+ [(mass of ice)(heat of fusion of ice)] = – [(mass of water)(specific heat of water)(change in temperature of water)]

Since

change in temperature of water = (final temperature of water) – (initial temperature of water)

Then the equation becomes

+ [(mass of ice)(heat of fusion of ice)] = – [(mass of water)(specific heat of water) {(final temperature of water) – (initial temperature of water)}]

Now, substituting the given values in the above equation we get: (Note that heat of fusion of ice is the same as the heat of fusion of water)

+ [(mass of ice)(6.01 kJ/mol)] = – [(64.3 g)(4.18 J/g•°C) {(15°C) – (55°C)}]

+ [(mass of ice)(6.01 kJ/mol)] = – [(64.3 g)(4.18 J/g•°C)(–40°C)]

We can also drop the outside brackets and drop the outside plus and minus signs:

(mass of ice)(6.01 kJ/mol) = (64.3 g)(4.18 J/g•°C)(40°C)

Since heat of fusion of ice is in kilojoules per mole, and specific heat of water is in joules per gram • degrees Celsius we need to convert kilojoules to joules.

In addition, since we need to find the amount of ice in grams, we need to convert moles of ice to grams as well. Therefore, the heat of fusion of ice should be expressed in joules per gram.

(mass of ice)(6,010 J/18.01528 g) = (64.3 g)(4.18 J/g•°C)(40°C)

or

(mass of ice)(333.6056 J/g) = (64.3 g)(4.18 J/g•°C)(40°C)

Dividing both sides by 333.6056 J/g we get:

mass of ice = [(64.3 g)(4.18 J/g•°C)(40°C)] / (333.6056 J/g)

mass of ice = 32.226557 g or 32 g rounded to two significant figures

Answer: The amount of ice at 0°C that was added to 64 grams of water at 55°C was about 32 grams.