a 100.0 gram metal sample is heated to a temperature of 110 degrees celcius then placed into a calorimeter containing 100.0 r of pure water (at initial temperature of 23 degrees celcius). after the sample and water reach thermal equalibrium, their final temperature is 29 degrees celsius. what is the specific heat capacity of the metal sample in calories/g*degrees celsius.

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To find the specific heat capacity of the metal sample, we can use the formula:

q = m * c * ΔT

Where:
q = heat energy absorbed or released by the substance
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature

In this case, the metal sample transfers heat to the water, resulting in a decrease in temperature for the metal and an increase in temperature for the water until they reach thermal equilibrium.

Let's calculate the values to substitute into the formula.

For the metal sample:
Mass (m) = 100.0 grams
Change in temperature (ΔT) = Final temperature - Initial temperature = 29°C - 110°C = -81°C (Note: We use a negative sign because the metal sample loses heat)
Heat energy (q) absorbed or released by the metal sample = m * c * ΔT

For the water:
Mass (m) = 100.0 g
Change in temperature (ΔT) = Final temperature - Initial temperature = 29°C - 23°C = 6°C
Heat energy (q) absorbed or released by the water = m * c * ΔT

Since the heat lost by the metal sample is transferred to the water, the heat energies for both are equal:

m_metal * c_metal * ΔT_metal = m_water * c_water * ΔT_water

Simplifying the equation, we get:

c_metal = (m_water * c_water * ΔT_water) / (m_metal * ΔT_metal)

Now we can substitute the values:

c_metal = (100.0 g * 1 calorie/g°C * 6°C) / (100.0 g * -81°C)

Solving the equation, the specific heat capacity of the metal sample is:

c_metal = -0.0741 calorie/g°C

Note that the negative sign indicates that the metal sample lost heat energy.