A 13.0 gram sample of an unknown metal at 30.0*C is placed in a Styrofoam cup containing 50.0 grams of water at 90.0*C.The water cools down and the metal warms up until thermal equilibrium is achieved at 87.0*C.Assuming all the heat lost by the water is gained by the metal and the cup is perfectly insulated,determine the specific of heat capacity of the unknown metal.The specific heat capacity of water is 4.18 J/g/*C
4.18 * (50.0/13.0) * [(90.0 - 87.0)/(90.0 - 30.0)]
To find the specific heat capacity of the unknown metal, we can use the equation:
q = m * c * ΔT
where:
q is the heat gained or lost (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in J/g°C), and
ΔT is the change in temperature (in °C).
In this case, we are assuming that all the heat lost by the water is gained by the metal. Therefore, the heat lost by the water equals the heat gained by the metal.
Let's calculate the heat lost by the water first using the equation above:
q water = m water * c water * ΔT water
Given:
m water = 50.0 g
ΔT water = (90.0 - 87.0) °C = 3.0 °C (since the water cools down from 90.0 °C to 87.0 °C)
Plugging in the values, we get:
q water = 50.0 g * 4.18 J/g°C * 3.0 °C
= 627 J
Since the heat lost by the water is equal to the heat gained by the metal, we have:
q metal = q water = 627 J
Next, we can calculate the mass of the metal using the equation:
q metal = m metal * c metal * ΔT metal
Given:
m metal = 13.0 g
ΔT metal = (87.0 - 30.0) °C = 57.0 °C (since the metal warms up from 30.0 °C to 87.0 °C)
Plugging in the values, we have:
627 J = 13.0 g * c metal * 57.0 °C
To find c metal, we rearrange the equation:
c metal = 627 J / (13.0 g * 57.0 °C)
≈ 0.87 J/g°C
Therefore, the specific heat capacity of the unknown metal is approximately 0.87 J/g°C.