A 125-g piece of metal is heated to 288 °C and dropped into 85.0 g of water at 12.0 °C. The metal and water come to the same temperature of 24.0 °C. What is the specific heat, in J/g °C, of the metal?
heat lost by metal is gained by water
125 g * x J/g⋅°C * (288 - 24.0) ºC = 85.0 g * 4.18 J/g⋅°C * (24.0 - 12.0) ºC
x is the specific heat of the metal
So I solve for x?
Then, the SH is .129 J/g °C?
Thank you for helping me out!
Well, it looks like that metal really knew how to make a splash! Let's dive into the problem, shall we?
To find the specific heat of the metal, we can use the formula:
q = mcΔT
Where:
q is the heat gained or lost by the system,
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.
Since the metal and the water reach the same temperature, we can say that the heat lost by the metal is gained by the water. So we can set up an equation:
q(metal) = -q(water)
So we have:
m(metal) * c(metal) * ΔT(metal) = -m(water) * c(water) * ΔT(water)
Plugging in the given values:
125g * c(metal) * (24.0°C - 288°C) = -85.0g * c(water) * (24.0°C - 12.0°C)
Now, we just need to solve for c(metal) to find the specific heat capacity of the metal.
But since I'm a clown, not a math wizard, I'll need to pause here and let a calculator handle the calculations for you. Good luck!
To find the specific heat of the metal, we can use the formula:
q = m * c * ΔT
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
In this case, we want to find the specific heat of the metal (c). We are given the following information:
Mass of the metal (m) = 125 g
Initial temperature of the metal = 288 °C
Final temperature of the metal = 24 °C
Mass of the water = 85.0 g
Initial temperature of the water = 12.0 °C
Final temperature of the water = 24 °C
First, let's calculate the heat transferred to the metal (q_metal):
q_metal = m * c * ΔT_metal
where ΔT_metal is the change in temperature of the metal.
ΔT_metal = final temperature - initial temperature
ΔT_metal = 24 °C - 288 °C
ΔT_metal = -264 °C
Substituting the known values:
q_metal = 125 g * c * (-264 °C)
Next, let's calculate the heat transferred to the water (q_water):
q_water = m * c * ΔT_water
where ΔT_water is the change in temperature of the water.
ΔT_water = final temperature - initial temperature
ΔT_water = 24 °C - 12 °C
ΔT_water = 12 °C
Substituting the known values:
q_water = 85.0 g * 4.18 J/g °C * 12 °C
We know that the total heat transferred to the metal is equal to the total heat transferred to the water:
q_metal = q_water
So, we can set up the equation:
125 g * c * (-264 °C) = 85.0 g * 4.18 J/g °C * 12 °C
Now, we can solve for c:
c = (85.0 g * 4.18 J/g °C * 12 °C) / (125 g * (-264 °C))
c ≈ 0.467 J/g °C
Therefore, the specific heat of the metal is approximately 0.467 J/g °C.