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kirza4 [7]
3 years ago
14

A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me

lts completely. If gold’s specific heat is 0.1291 joules/gram degree Celsius and its heat of fusion is 63.5 joules/gram, how much energy is gained by the gold?
The gold gains a total of______joules of energy.


A kettle of water is at 14.5°C. Its temperature is then raised to 50.0°C by supplying it with 5,680 joules of heat. The specific heat capacity of water is 4.186 joules/gram degree Celsius. What is the mass of water in the kettle? Express your answer to three significant figures.

The mass of the water in the kettle is______ grams.
Chemistry
2 answers:
DiKsa [7]3 years ago
8 0

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

g100num [7]3 years ago
4 0

Answer:

2480 J; 38.2 g

Explanation

<em>Question 1 </em>

There are two heat flows in this problem

<em>q</em> = Heat to raise gold to its melting point + heat to melt the gold

<em>q</em> =                              <em>q</em>₁                                +               <em>q</em>₂

<em>q</em> =                           mcΔT                             +            mΔH

===============

<em>Heat to heat the gold (q₁) </em>

Δ<em>T</em> = T_<em>f</em> – T_<em>i</em>                      Insert the values

ΔT = 1064 – 26                    Do the subtraction

ΔT= 1038 °C                         Insert values into the formula for q₁

q₁ = 12.4 × 0.1291 × 1038     Do the multiplication

q₁ = 1662 J

==============

<em>Heat to melt the gold (q₂) </em>

<em>q</em>₂ = 12.4 × 63.5                    Do the multiplication

<em>q</em>₂ = 787.4 J

===============

<em>Total heat required (q) </em>

<em>q</em> = 1662 + 787.4                   Do the addition

<em>q</em> = 2450 J  (to three significant figures)

===============

<em>Question 2</em>

<em>q = mC</em>Δ<em>T</em>

Δ<em>T</em> = <em>T</em>_f – <em>T</em>_i                       Insert the values

Δ<em>T</em> = 50.0 – 14.5                   Do the subtraction

Δ<em>T</em>= 35.5 °C                          Insert values into the formula for <em>q</em>

5680 = <em>m</em> × 4.186 × 35.5      Do the multiplication

5680 = <em>m</em> × 148.6                  Divide both sides by 148.6

5680/148.6 = <em>m</em>                     Do the division and switch

<em>m</em> = 38.2 g

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