An insulated beaker contains 250.0 grams of water at 25oC. Exactly 41.6 g of a
metal at 100.0oC was dropped in the beaker. The final temperature of the water
was 26.4oC. Assuming that no heat is lost in any other way, calculate the specific heat of the metal.
a. 0.159 J g-1 K-1
b. 0.478 J g-1 K-1
c. 0.0503 J g-1 K-1
d. 2.09 J g-1 K-1
i have no idea how to do this?
I did this for Bob about a half hour ago. Here is a link.
http://www.jiskha.com/display.cgi?id=1327199259
To solve this question, we can use the principle of conservation of energy. According to this principle, the heat lost by the metal is equal to the heat gained by the water.
The equation for heat transfer is given by:
Q = m * c * ΔT
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*K)
ΔT is the change in temperature (in Kelvin)
In this case, the heat lost by the metal is equal to the heat gained by the water. So we can set up the equation:
m * c * ΔT_metal = m * c * ΔT_water
Where:
ΔT_metal is the change in temperature for the metal (final temperature - initial temperature of the metal)
ΔT_water is the change in temperature for the water (final temperature - initial temperature of the water)
Plugging in the given values, we have:
41.6g * c * (26.4°C - 100.0°C) = 250.0g * 4.18 J/g°C * (26.4°C - 25°C)
Simplifying, we get:
-4138.24c = 41.5
Dividing both sides by -4138.24, we find:
c ≈ 0.01 J/g°C
Since the given options are in J/g*K, we need to convert °C to K by adding 273.15:
c ≈ 0.01 J/g*K
Therefore, the specific heat of the metal is approximately 0.01 J/g*K.
Since none of the given options directly match this value, we can select the closest option, which is:
c. 0.0503 J/g*K
Hence, the specific heat of the metal is approximately 0.0503 J/g*K.