A hot, just minted copper coin is placed in 112g of water to cool. The water temperature changes by 8.58 C and the temperature of the coin changes by 80.0 C. What is the mass of the coin? Disregard any energy trans to the waters surroundings and assume the specific heat of copper is 387 j/kg C. The heat of water is 4186 J/kg C. Answer in units of g.
To solve this problem, we need to calculate the heat energy transferred from the copper coin to the water using the equation:
Q = mcΔT
Where:
Q = heat energy transferred
m = mass
c = specific heat capacity
ΔT = change in temperature
First, let's calculate the heat energy transferred from the water:
Qwater = mwater * cwater * ΔTwater
Qwater = 112g * 4186 J/kg°C * 8.58°C
Qwater = 417752.16 J
Next, let's calculate the mass of the copper coin:
Qcopper = mcopper * ccopper * ΔTcopper
We can rearrange the equation to solve for mcopper:
mcopper = Qcopper / (ccopper * ΔTcopper)
mcopper = 417752.16 J / (387 J/kg°C * 80.0°C)
mcopper ≈ 13.71 kg
Lastly, we need to convert the mass from kg to g:
mcopper = 13.71 kg * 1000 g/kg
mcopper ≈ 13710 g
Therefore, the mass of the copper coin is approximately 13710 g.
To solve this problem, we will use the principle of conservation of energy. The heat lost by the hot copper coin will be gained by the water as it cools down.
The amount of heat transferred can be calculated using the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred,
m is the mass of the substance (coin or water),
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
Let's start with the heat transferred to the water:
Q_water = m_water * c_water * ΔT_water
Substituting the given values:
ΔT_water = 8.58 °C (given)
c_water = 4186 J/kg °C (specific heat of water, given)
We are given the mass of the water, which is 112g. However, we need to convert it to kilograms to use in the equation:
m_water = 112 g = 0.112 kg
Now we can calculate the heat transferred to the water:
Q_water = m_water * c_water * ΔT_water
= 0.112 kg * 4186 J/kg °C * 8.58 °C
= 3967.7 J
Now, let's calculate the heat transferred to the copper coin:
Q_coin = m_coin * c_coin * ΔT_coin
We are looking for the mass of the coin, so let’s solve for it:
m_coin = Q_coin / (c_coin * ΔT_coin)
Substituting the given values:
ΔT_coin = 80.0 °C (given)
c_coin = 387 J/kg °C (specific heat of copper, given)
We know that the heat transferred to the water is equal to the heat transferred to the coin:
Q_coin = Q_water = 3967.7 J
Now we can calculate the mass of the coin:
m_coin = Q_coin / (c_coin * ΔT_coin)
= 3967.7 J / (387 J/kg °C * 80.0 °C)
= 0.128 kg
Finally, we need to convert the mass of the coin from kilograms to grams:
m_coin = 0.128 kg = 128 g
Therefore, the mass of the coin is 128 g.
Equate the heat loss of the coin and the heat gain of the water.
Heat transfer is M*C*(delta T)
Mcoin*387*80 = 0.112*4186*8.58
The only unknown is Mcoin.
Solve for it.