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kondor19780726
11 days ago
10

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water. Remember, the density of water is 1.0 g/m
Physics
1 answer:
ValentinkaMS [1.1K]11 days ago
7 0

Answer:

335°C

Explanation:

Heat that is either absorbed or released can be expressed as:

q = m C ΔT

where m represents mass, C signifies specific heat capacity, and ΔT indicates temperature change.

The heat gained by water equals the heat lost by copper.

mw Cw ΔTw = mc Cc ΔTc

The final temperature for both water and copper is the same, thus:

mw Cw (T - Tw) = mc Cc (Tc - T)

Provided:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Required: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

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The final mass will be slightly lower due to evaporation. I learned this back in third grade, so it's surprising you're in high school and don't know this.
5 0
7 days ago
A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108
Ostrovityanka [942]

Answer: The calorimeter's heat capacity is 6.72J/g^oC

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat capacity of water = 4.184J/g^oC

c_2 = specific heat capacity of calorimeter =?

m_1 = mass of water = 108.7 g

m_2 = mass of calorimeter = 108.7 g

T_f = final temperature of the mixture = 35.0^oC

T_1 = initial temperature of the water = 60.2^oC

T_2 = initial temperature of calorimeter = 19.3^oC

Now substituting all provided values into the formula, we obtain

(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC

c_2=6.72J/g^oC

Hence, the heat capacity of the calorimeter is 6.72J/g^oC

3 0
9 days ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [1105]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock  

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

 (m)child(d)child)=(m)rock(d)rock)  

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0
5 days ago
A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
Keith_Richards [1034]

Response:

Q = 8,345 * v

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Q = 8,345 * v

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

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3 days ago
By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a
Ostrovityanka [942]

Broad questions addressed by conducting this experiment involve the effects of electric current.

Additional details

Electric current measures the quantity of electric charge passing per unit time.

It results from electrons moving due to a voltage difference (high potential to low potential) between two points.

These electrons flow through wires acting as conductors.

Ohm's Law states that:

The potential difference across a conductor is proportional to the current flowing through it, assuming resistance remains the same.

\displaystyle I=\frac{V}{R}\\\\V=I\times R

A basic electrical circuit consists of a voltage source (battery) and a lamp.

Ammeters used to measure current must be connected in series with the load.

By adjusting the voltage while resistance is constant, varying current values are observed; increasing voltage produces higher current.

Learn more

Electron flow inside devices

brainly.com/question/4438943

Keywords: basic electric circuits, Ohm's law, experiment

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13 days ago
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