Question

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 melts 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.

Answer 1

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.

Answer 2

Answer:

2480 J; 38.2 g

Explanation

Question 1

There are two heat flows in this problem

q = Heat to raise gold to its melting point + heat to melt the gold

q =                              q₁                                +               q₂

q =                           mcΔT                             +            mΔH

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

Heat to heat the gold (q₁)

ΔT = T_f – T_i                      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

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

Heat to melt the gold (q₂)

q₂ = 12.4 × 63.5                    Do the multiplication

q₂ = 787.4 J

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

Total heat required (q)

q = 1662 + 787.4                   Do the addition

q = 2450 J  (to three significant figures)

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

Question 2

q = mCΔT

ΔT = T_f – T_i                       Insert the values

ΔT = 50.0 – 14.5                   Do the subtraction

ΔT= 35.5 °C                          Insert values into the formula for q

5680 = m × 4.186 × 35.5      Do the multiplication

5680 = m × 148.6                  Divide both sides by 148.6

5680/148.6 = m                     Do the division and switch

m = 38.2 g