Question

16.0 J of heat is added to 10.00 g of gold, which is initially at 25°C.

The heat capacity of gold is 25.41 J/(mol • °C).

What would be the final temperature of the gold?

Answer 1

The final temperature of the gold is 37.4°C.

Step-by-step explanation:

First, we need to calculate the number of moles of gold:

`\sf\dashrightarrow Moles = (Mass)/(Molar\: mass)`

The molar mass of gold is 196.9665 g/mol, so:

`\sf\dashrightarrow Moles = (10.00\: g)/(196.9665\: g/mol) = \bold{0.05078\: mol}`

Next, we can use the formula for heat capacity to calculate the heat absorbed by the gold:

`\sf\qquad\dashrightarrow \Delta Q = m * C * \Delta T`

where:

• ΔQ is the heat absorbed
• m is the mass of the gold
• C is the heat capacity of gold
• ΔT is the change in temperature

Rearranging this equation to solve for the change in temperature, we get:

`\sf\dashrightarrow \Delta T = (\Delta Q)/((m * C))`

Plugging in the given values, we get:

`\sf\dashrightarrow\Delta T = \frac{16.0\: J}{(0.05078\: mol * 25.41\: J/(mol \cdot {}^(\circ)C))}`

`\sf\dashrightarrow\Delta T = 12.4^(\circ)C`

Therefore, the final temperature of the gold would be:

`\sf\qquad\dashrightarrow T_f = T_i + \Delta T`

`\sf\qquad\dashrightarrow T_f = 25^(\circ)C + 12.4^(\circ)C`

`\sf\qquad\dashrightarrow T_f = \boxed{\bold{\:\:37.4^(\circ)C\:\:}}`

Answer 2

`T_(2)=`37.40°C.

Step-by-step explanation:

1. Theory.

The amount of heat gained or lost by a determined amount of mass of any substance can be calculated using the equation q = mcΔT; where "q" is the heat added or subtracted from the mass; "m" is the value of the mass; "c" is the heat capacity of the substance, and "ΔT" is the change in temperature before and after applying or removing the energy.

2. Identify the information based on the formula.

q= 16.0 J

m (moles)= m(mass)/M(molar mass)

Looking up the molar mass on the periodic table, we can see that it equals 196.96657 u, which is equal to 196.96657 g/mole.

m (moles)= (10.00g)/(196.96657 g/mole)= 0.0508 moles.

c= 25.41 J/(mol • °C).

ΔT=
`T_(2) -T_(1)`

`T_(1)`= 25°C

`T_(2)`= ?

3. Rewrite the formula.

q = mcΔT

Since ΔT=
`T_(2) -T_(1)`, then →
`q = mc(T_(2) -T_(1) )`

4. Solve the formula for
`T_(2)`.

`q = mc(T_(2) -T_(1) )\\\\ (q)/(mc) = (T_(2) -T_(1) )\\\\(q)/(mc) +T_(1)= T_(2)\\ \\T_(2)=(q)/(mc) +T_(1)`

5. Calculate
`T_(2)`.

`T_(2)=(16.0J)/((0.0508moles)(25.41(J)/(mol*C) )) +25C`

`T_(2)=`37.40°C.