Problem with calculating final temperature of water

[v3ra]

A piece of ice with a mass of 60 g is transferred directly from the freezer at a
temperature of –7°C to a polystyrene cup containing 200 g of water at 75°C. Calculate
the final temperature of the water. Ignore any heat loss to the cup and its surroundings.

Qheat ice + Qmelt ice + Qheat cold water + Qcool hot water
mciceΔtice + mLf + mcwΔt + mhwcwΔthw = 0
(0.060 kg) x (2100 J/(kg.C)) x (0°C - (-7°C)) + ( 0.060 kg) x (3.3 x 10^5) + (0.060 kg) x (4200 J /(kg.C)) x (t2 – 0°C) + (0.2) x (4200 J) x (t2 – 75°C) = 0
882 + 19800 + 252t2 + 840t2 – 63000 = 0
840t2 = 42318
t2 = 50°C

The above problem is from my book, but I am having difficulty with the equation. Where did
the 63000 come from?

 

[TSny]

A piece of ice with a mass of 60 g is transferred directly from the freezer at a
temperature of –7°C to a polystyrene cup containing 200 g of water at 75°C. Calculate
the final temperature of the water. Ignore any heat loss to the cup and its surroundings.

Qheat ice + Qmelt ice + Qheat cold water + Qcool hot water
mciceΔtice + mLf + mcwΔt + mhwcwΔthw = 0
(0.060 kg) x (2100 J/(kg.C)) x (0°C - (-7°C)) + ( 0.060 kg) x (3.3 x 10^5) + (0.060 kg) x (4200 J /(kg.C)) x (t2 – 0°C) + (0.2) x (4200 J) x (t2 – 75°C) = 0
882 + 19800 + 252t2 + 840t2 – 63000 = 0
840t2 = 42318
t2 = 50°C

The above problem is from my book, but I am having difficulty with the equation. Where did
the 63000 come from?

Hi, V3ra. Multiply the three red numbers.

 

[v3ra]

Hmm, but how would I have known to do that? I already multiplied the 0.2 and 4200 J to get 840. Where in the equation is it implied that I would use these two numbers again along with the 75°C to get 63000?

 

[frogjg2003]

It's called the distributive property.
a*(b+c)=a*b+a*c
You have a*(b+c)=a*b

 

[v3ra]

Can you please elaborate? How would someone who is approaching the above equation for the first time know to apply such a rule? The first part of the equation contains three numbers that are being multiplied, would all three be represented by "a" in the "a*(b+c)=a*b" or is this rule only applying to one part of the equation?

 

[TSny]

Can you please elaborate? How would someone who is approaching the above equation for the first time know to apply such a rule? The first part of the equation contains three numbers that are being multiplied, would all three be represented by "a" in the "a*(b+c)=a*b" or is this rule only applying to one part of the equation?

v3ra, I'm not sure of your question. You are trying to simplify (0.2) x (4200) x (t₂ – 75). You noted earlier that you can multiply the 0.2 and the 4200 to get 840. So now you have 840 x (t₂-75).

To reduce this, take a simple example: 3 x (6 - 2). This means you have the quantity (6-2) three times.
So, 3 x (6 - 2) = (6-2) + (6-2) + (6-2). So, you get "6" three times and you also get (-2) three times.

So, 3 x (6-2) = 3x(6) + 3x(-2) = 3x6 - 3x2. Thus 3 x (6-2) = 3x6 - 3x2

Likewise, 840 x (t₂-75) means that you have 840 of the quantity (t₂-75). That's 840 t₂ 's and also 840 of the quantity (-75).

So, 840 x (t₂-75) = 840 x t₂ + 840 x (-75) = 840 x t₂ - 840x75 = 840t₂ - 63000.

 

[v3ra]

I understand now. I was, for some reason, completely avoiding the "t2".

 

[frogjg2003]

No, you had t2, it was the -75 that you left out. That's the term that became the -63000.