In the laboratory a \"coffee cup\" calorimeter , or constant pressure calorimete

Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction.

A student heats 69.66 grams of platinum to 98.86 °C and then drops it into a cup containing 84.34 grams of water at 21.32 °C. She measures the final temperature to be 23.24 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.67 J/°C.

Assuming that no heat is lost to the surroundings calculate the specific heat of platinum.

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction.

A student heats 69.66 grams of platinum to 98.86 °C and then drops it into a cup containing 84.34 grams of water at 21.32 °C. She measures the final temperature to be 23.24 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.67 J/°C.

Assuming that no heat is lost to the surroundings calculate the specific heat of platinum.

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific Thermometer heat of a solid, or to measure the energy of a solution phase reaction Stirring rod A student heats 69.66 grams of platinum to 98.86 °C and then drops it into a cup containing 84.34 grams of water at 21.32 °C. She measures the final temperature to be 23.24 °C The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.67 J/oC Assuming that no heat is lost to the surroundings calculate the specific heat of platinum Specific Heat (Pt) = Water Metal sample J/g C.

Explanation / Answer

Solution:- Water temperature is increasing it means water gained energy from the metal.

change in temperature, delta T for water = 23.24 - 21.32 = 1.92 degree C

mass of water = 84.34 g

specific heat of water = 4.184 J/g. degree C

heat gained by water is calculated by using the formula,

q = m c delta T

q = 84.34 g x 4.184 J/g.degree C x 1.92 degree C

q = 677.53 J

heat gained by calorimeter is calculated by using the equation,

q = Cp*delta T

where Cp is the heat capacity of calorimeter.

q = 1.67 J/degree C x 1.92 degree C

q = 3.21 J

Heta lost by metal = - (heat gained by water + heat gained by calorimeter)

heat lost by metal = - (677.53 J + 3.21 J)

heat lost by metal = -680.74 J

delta T for metal = 23.24 - 98.86 = -75.62 degree C

mass of metal, m = 69.66 g

we would use the equation, q = m c delta T to calculate the specific heat of metal.

c = q/m*delta T

c = -680.74 J/(-69.66 g * 75.62 degree C

c = 0.129 J/g.degree C

So, the specific heat of the metal(Pt) is 0.129 J/g. degree C.

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