If 75.0 g of metal at 75.0 c is added to 135 g of water at 15 c the temp of water rises to 18.3 c what is the specific heat of the metal..?
Can someone please help me plug the #s into the specific heat formula I'm very confused as to were to plug these #s in
I need to understand your confusion.
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] =0
It simply makes no sense to me that you can't see mass metal is 75.0 g, etc.
I'm confused when it comes to the temperatures.... I don't get as to why you're here trying to call me stupid when I just need some help.
The answer I got is .1047 cal/ g c
Did I call you stupid? NO, I didn't. If you want help you must help us help you by telling us what you don't understand. Telling me you can't tell 75.0 g is mass metal and 135 g is mass H2O is ridiculous. What you're trying to do is have us work the problem for you.
OK, so temperature. Tfinal is the temperature after the system reaches equilibrium and will be the same for the metal and water. Tinitial is the beginning temperature of each.
I figured it out already thanks
Certainly! I can help you with that.
The specific heat formula is given by:
Q = m * c * ΔT
Where:
Q is the heat transferred (in joules),
m is the mass (in grams),
c is the specific heat (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
In your case:
- The mass of the metal (m) is 75.0 g.
- The initial temperature of the metal (Ti) is 75.0 °C.
- The mass of the water (m) is 135 g.
- The initial temperature of the water (Ti) is 15 °C.
- The final temperature of the water (Tf) is 18.3 °C.
Now, let's plug these values into the formula:
Q = m * c * ΔT
The heat transferred (Q) can be calculated by:
Q = mass of metal * specific heat of metal * change in temperature of metal
m(Metal) = 75.0 g
c(Metal) = unknown (specific heat of the metal, what we're trying to find)
ΔT(Metal) = Tf(Water) - Ti(Water) = 18.3 °C - 15 °C = 3.3 °C
Q = 75.0 g * c(Metal) * 3.3 °C
On the other hand, we know that the heat transferred to the water is equal to the heat transferred from the metal. So we can express this as:
Q(Metal) = Q(Water)
Now let's calculate the heat transferred to the water:
Q(Water) = m(Water) * c(Water) * ΔT(Water)
m(Water) = 135 g
c(Water) = 4.18 J/g°C (specific heat of water)
ΔT(Water) = Tf(Water) - Ti(Water) = 18.3 °C - 15 °C = 3.3 °C
Q(Water) = 135 g * 4.18 J/g°C * 3.3 °C
Since Q(Metal) = Q(Water), we can set these two equations equal to each other and solve for c(Metal):
75.0 g * c(Metal) * 3.3 °C = 135 g * 4.18 J/g°C * 3.3 °C
Now we can solve for c(Metal) by dividing both sides of the equation by (75.0 g * 3.3 °C):
c(Metal) = (135 g * 4.18 J/g°C * 3.3 °C) / (75.0 g * 3.3 °C)
By doing the calculation, you'll find the specific heat of the metal.