Agoldsmith 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 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 grams.

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.

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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2