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A 25.0 g sample of warm water at 40.0⁰C was added to a 25.0 g sample of water in a Styrofoam coffee cup calorimeter initially at 20.0⁰C. The final temperature of the mixed water and calorimeter was 29.5⁰C. Calculate the heat capacity of the coffee cup calorimeter. The specific heat of water is 4.184 J/g∙⁰C.

a.
0.189 J/⁰C

b.
27.3 J/⁰C

c.
11.0 J/⁰C

d.
116 J/⁰C

User Itzik
by
7.7k points

2 Answers

4 votes

Final answer:

The heat capacity of the coffee cup calorimeter is calculated by using the formula q = m · c · ΔT with the principle of conservation of energy and accounting for heat lost by warm water and gained by cooler water.

Step-by-step explanation:

To find the heat capacity of the coffee cup calorimeter, we can use the concept of conservation of energy, which implies that heat lost by the warm water will be gained by the cooler water and the calorimeter.

Since the specific heat of water is given as 4.184 J/g°C, we can calculate the heat exchanged using the formula

q = m · c · ΔT,

where q is the heat exchange,

m is the mass,

c is the specific heat capacity,

and ΔT is the change in temperature.

Both water samples have the same mass, and thus each will lose/gain the same amount of heat to reach the final temperature.

Therefore, we calculate the heat change for the warm water: q₁ = m · c · (ΔT) = 25.0 g · 4.184 J/g°C · (29.5°C - 40.0°C).

Since warm water is cooling down, the ΔT will be negative, indicating heat is lost.

Likewise, the cooler water's heat change is q₂ = m · c · (ΔT) = 25.0 g ·

4.184 J/g°C · (29.5°C - 20.0°C).

A positive ΔT indicates heat is gained. The calorimeter gains the remaining heat.

Assuming no heat loss to the surroundings, q₁ + q₂ + q₃(calorimeter) = 0. Solving for q₃(calorimeter) gives us the heat capacity of the coffee cup calorimeter, which can be calculated and matched to the options provided.

User Boss COTIGA
by
8.0k points
2 votes

Answer:

2024.70 J

Step-by-step explanation:

The heat capacity of the coffee cup calorimeter can be calculated using the following formula:

q_calorimeter = q_water + q_water_final

where q_calorimeter is the heat absorbed by the coffee cup calorimeter, q_water is the heat lost by the warm water, and q_water_final is the heat gained by the cold water.

First, calculate q_water:

q_water = m_water * c_water * ΔT

where m_water = 25.0 g is the mass of the warm water, c_water = 4.184 J/g°C is the specific heat of water, and ΔT = (40.0°C - 29.5°C) = 10.5°C is the change in temperature.

q_water = 25.0 g * 4.184 J/g°C * 10.5°C = 1057.35 J

Next, calculate q_water_final:

q_water_final = m_water * c_water * ΔT

where m_water = 25.0 g is the mass of the cold water, c_water = 4.184 J/g°C is the specific heat of water, and ΔT = (29.5°C - 20.0°C) = 9.5°C is the change in temperature.

q_water_final = 25.0 g * 4.184 J/g°C * 9.5°C = 967.35 J

Finally, calculate the heat capacity of the coffee cup calorimeter:

q_calorimeter = q_water + q_water_final = 1057.35 J + 967.35 J = 2024.70 J

So the heat capacity of the coffee cup calorimeter is 2024.70 J.

User Misterbassman
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8.1k points