When 45 grams of an alloy is dropped into 100.0 grams of water at 25 degrees celcius, the final temperature is 37 degrees celcius. what is the specific heat of the alloy? the alloys Temperature is 100 degrees celsius
heat lost by alloy + heat gained by water = 0
heat lost or gained = mass x specific heat x (Tfinal-Tinitial).
mass alloy x specific heat alloy x (Tfinal-Tinitial) + mass water x specific heat water x (Tfinal-Tinitial) = 0
You must look up the specific heat of water, the specific heat of the alloy is the only unknown, you know the mass of water and mass of the alloy. I note you spelled Celsius two ways, the correct way is Celsius, not Celcius, probably just a typo. Post your work if you get stuck.
To determine the specific heat of the alloy, we can use the heat equation, which is:
Q = mcΔT
Where:
Q is the heat transferred (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (final temperature - initial temperature), in this case, it is (37°C - 25°C)
First, let's calculate the heat transferred by the water.
Q_water = m_water * c_water * ΔT_water
Given:
m_water = 100.0 g (mass of water)
c_water = 4.18 J/g·°C (specific heat capacity of water)
ΔT_water = (37°C - 25°C) = 12°C
Q_water = (100.0 g) * (4.18 J/g·°C) * (12°C)
Q_water = 5025.6 J
Since the heat gained by the water must equal the heat lost by the alloy, we can set up the equation:
Q_water = Q_alloy
5025.6 J = m_alloy * c_alloy * ΔT_alloy
Given:
m_alloy = 45 g (mass of the alloy)
ΔT_alloy = (37°C - 100°C) = -63°C (change in temperature of the alloy since it started at 100°C)
c_alloy = ?
Now, we can rearrange the equation to solve for the specific heat capacity of the alloy (c_alloy):
c_alloy = Q_alloy / (m_alloy * ΔT_alloy)
c_alloy = 5025.6 J / (45 g * -63°C)
c_alloy ≈ -1.41 J/g·°C
The specific heat of the alloy is approximately -1.41 J/g·°C.