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Please help 30 pts!!!!! A friend needs these answers and I don't know what to say. If you don't know the answer, please ask somebody who might know, and tell @TracySparkle1

A student heats a 15.0 gram metallic sphere of unknown composition to a temperature of 98°C. The sphere is transferred to a calorimeter containing 100. mL of water at a temperature of 25.0°C. The student observes that the resulting temperture of both the water and the object is 27.1°C after the object is submerged.

1. Describe, in terms of the object and the water, the flow of heat energy that took place during the experiment.

2. Calculate the amount of heat energy gained by the water in the calorimeter.
Q = mc∆T???

3. Using the quantity of heat calculated in the previous question, determine the specific heat of the object.

Respuesta :

barnuts

1. Heat always flow from a region of higher temperature to a lower temperature, just like concentration gradient. So in this case, since the metallic sphere is hotter than water, so naturally heat energy flows from the metallic sphere to the water.

 

 

2. We use the formula:

Q = m C ∆T

where Q is the heat energy gained, m is mass of water, C is heat capacity of water = 4.184 J/g°C and ΔT is change in temperature

 

Since density of water is 1 g/mL, so mass is also 100g, therefore:

 

Q = 100 g * 4.184 J/g°C * (27.1°C - 25°C)

 

Q = 878.64 J

 

 

3. Heat gained by the water is equal to the heat lost by the metal, therefore:

Q(water) = - Q(metal)

878.64 J = - 15 g * C * (27.1°C - 98°C)

 

C = 0.83 J/g°C

jonathanrickyhiers

Answer:

1. Heat always flow from a region of higher temperature to a lower temperature, just like concentration gradient. So in this case, since the metallic sphere is hotter than water, so naturally heat energy flows from the metallic sphere to the water.

 

 

2. We use the formula:

Q = m C ∆T

where Q is the heat energy gained, m is mass of water, C is heat capacity of water = 4.184 J/g°C and ΔT is change in temperature

 

Since density of water is 1 g/mL, so mass is also 100g, therefore:

 

Q = 100 g * 4.184 J/g°C * (27.1°C - 25°C)

 

Q = 878.64 J

 

 

3. Heat gained by the water is equal to the heat lost by the metal, therefore:

Q(water) = - Q(metal)

878.64 J = - 15 g * C * (27.1°C - 98°C)

 

C = 0.83 J/g°C

Explanation:

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