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 27.5 g of water at T = 17.0°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 )

Well, this coin dealer sure knows how to make a hot investment! Let's calculate the final temperature and find out if it's a real silver surprise or a counterfeit nickel.

First, let's find the heat absorbed by the coin. We can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat absorbed by the coin can be calculated as follows: Q_coin = mcΔT = 14.0 g × 0.233 J/g°C × (100°C - 17.0°C).

Now, let's calculate the heat released by the coin to the water. Since the heat lost by the coin is equal to the heat gained by the water, we have: Q_water = mwater × cwater × ΔT.

Substituting the given values, we get: Q_water = 27.5 g × 4.18 J/g°C × (T - 17.0°C), where T is the final temperature of the water.

Since the heat absorbed by the coin and the heat released by the water are equal, we can equate the two expressions and solve for T.

14.0 g × 0.233 J/g°C × (100°C - 17.0°C) = 27.5 g × 4.18 J/g°C × (T - 17.0°C).

Solving this equation will give us the final temperature of the water.

Now, I know this calculation might seem a bit dry, but trust me, once you flip that coin and reveal its true identity, it will be more thrilling than any magic trick. Good luck, coin dealer!

To solve this problem, we will use the principle of heat transfer and the equation:

q = m * s * ΔT

where:
q = heat transferred (in J)
m = mass (in g)
s = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)

Let's calculate the heat transferred between the heated coin and the water:

1. Heat transferred from the coin to the water:

q1 = m1 * s1 * ΔT1
= 14.0 g * 0.233 J/g°C * (100°C - T)
= 3.262 g * (100°C - T) J

2. Heat transferred from the water to raise its temperature:

q2 = m2 * s2 * ΔT2
= 27.5 g * 4.45 J/g°C * (T - 17.0°C)
= 122.375 g * (T - 17.0°C) J

Since heat transfer is conserved, we can equate q1 and q2:

3.262 g * (100°C - T) J = 122.375 g * (T - 17.0°C) J

Now, let's solve this equation for T:

3.262 * 100 - 3.262 * T = 122.375 * T - 122.375 * 17.0
326.2 - 3.262T = 122.375T - 2081.875
326.2 + 2081.875 = 3.262T + 122.375T
2408.075 = 125.637T
T = 2408.075 / 125.637
T ≈ 19.150°C

Therefore, the final temperature of the water would be approximately 19.150°C.

To solve this problem, we need to use the concept of heat transfer. The heat gained by the water is equal to the heat lost by the coin. We can calculate the heat gained by the water using the formula:

Q = m * s * ΔT

Where:
Q is the heat gained or lost (in Joules)
m is the mass of the water (in grams)
s is the specific heat capacity of water (in J/g°C)
ΔT is the change in temperature (in °C)

Since the water gains heat from the coin, the heat lost by the coin will be equal in magnitude but opposite in sign.

First, let's calculate the heat lost by the coin. We know that the initial temperature was 100°C, and the final temperature is the final temperature of the water.

Q_lost = m_coin * s_silver * ΔT

Where:
m_coin is the mass of the coin (in grams)
s_silver is the specific heat capacity of silver (in J/g°C)
ΔT is the change in temperature (in °C)

Since the coin loses heat to the water, the change in temperature is given by:

ΔT = T_final - T_initial (water temperature)

Now, we can set up the equation equating the heat lost by the coin to the heat gained by the water:

Q_lost = -Q_gained

m_coin * s_silver * ΔT = -(m_water * s_water * ΔT)

Since ΔT is common to both sides of the equation, we can simplify the equation to:

m_coin * s_silver = -m_water * s_water

Now we can solve for the final temperature of the water (T_final):

T_final = T_initial + (m_coin * s_silver) / (m_water * s_water)

Plugging in the given values:

T_initial = 17.0°C
m_coin = 14.0 g
s_silver = 0.233 J/g°C
m_water = 27.5 g
s_water = 4.18 J/g°C (specific heat capacity of water)

T_final = 17.0 + (14.0 * 0.233) / (27.5 * 4.18)

Calculating this value, we find:

T_final ≈ 17.0 + 0.164 = 17.164°C

Therefore, the final temperature of the water would be approximately 17.164°C.

heat lost by coin + heat gained by H2O = 0

[(mass coin x sp.h Ag x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tfinal if it is Ag