A coin dealer, offered a rare silver coin, suspected that it might be a counterfeit nickel copy. The dealer heated the coin, which weighed 14.0 g to 100°C in boiling water and then dropped the hot coin into 29.5 g of water at T = 15.0°C in an insulated coffee-cup, and measured the rise in temperature. If the coin was really made of silver, what would the final temperature of the water be (in °C)? (for nickel, s = 0.445 J/g°C; for silver, s = 0.233 J/g°C )

Question

A coin dealer, offered a rare silver coin, suspected that it might be a counterfeit nickel copy. The dealer heated the coin, which weighed 14.0 g to 100°C in boiling water and then dropped the hot coin into 29.5 g of water at T = 15.0°C in an insulated coffee-cup, and measured the rise in temperature. If the coin was really made of silver, what would the final temperature of the water be (in °C)? (for nickel, s = 0.445 J/g°C; for silver, s = 0.233 J/g°C )

Answer 1

To solve this problem, we can use the principle of conservation of energy, specifically the heat transfer equation:

Q = mcΔT

Where:
- Q is the heat transferred
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature

We can calculate the heat transferred to the water and the coin separately, and then set them equal to each other since they both reach the same final temperature.

1. Calculate the heat transferred to the water:
Q_water = mw * cw * ΔTw
Where:
- mw is the mass of the water
- cw is the specific heat capacity of water
- ΔTw is the change in temperature

Given values:
mw = 29.5 g
cw = 4.18 J/g°C (specific heat capacity of water)
ΔTw = Tf - T_i = Tf - 15.0°C (final temperature - initial temperature)

2. Calculate the heat transferred to the coin:
Q_coin = mc * cc * ΔTc
Where:
- mc is the mass of the coin (14.0 g)
- cc is the specific heat capacity of the coin (unknown for now)
- ΔTc is the change in temperature (100°C - Tf)

Now, since the heat transfer between the two objects is equal to each other:
Q_water = Q_coin

Substituting the equations:
mw * cw * ΔTw = mc * cc * ΔTc

Now, solve for Tf, the final temperature of the water.

1. Rearrange the equation:
mw * cw * ΔTw = mc * cc * ΔTc
mw * cw * (Tf - 15.0°C) = mc * cc * (100°C - Tf)

2. Distribute and collect like terms:
mw * cw * Tf - mw * cw * 15.0°C = mc * cc * 100°C - mc * cc * Tf

3. Collect the terms with Tf on one side:
(mw * cw + mc * cc) * Tf = mw * cw * 15.0°C + mc * cc * 100°C

4. Solve for Tf:
Tf = (mw * cw * 15.0°C + mc * cc * 100°C) / (mw * cw + mc * cc)

Now, plug in the values and calculate Tf using the given specific heat capacities for nickel and silver (s).

For nickel:
cc = 0.445 J/g°C

For silver:
cc = 0.233 J/g°C

Plug the appropriate values into the equation to find the final temperature (Tf) of the water.