A248-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

The initial temperature of the piece of copper is 335.6 °C. Step 1:

Data obtained from the question

Mass of copper (M꜀) = 248 g

Volume of water  = 390 mL

Density of water = 1 g/mL

Initial temperature of water (Tᵥᵥ) = 22.6 °C

Equilibrium temperature (Tₑ) = 39.9 °C

Initial temperature of copper (T꜀) =? Step 2:

Determination of the mass of water

Volume of water = 390 mL

Density of water = 1 g/mL

Mass of water =?

Cross multiply

Mass of water = 390 gStep 3:

Determination the initial temperature of the copper.

Mass of copper (M꜀) = 248 g

Mass of water (Mᵥᵥ) = 390 g

Initial temperature of water (Tᵥᵥ) = 22.6 °C

Equilibrium temperature (Tₑ) = 39.9 °C

Initial temperature of copper (T꜀) =? NOTE:

1. Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

2. Specific heat capacity of copper (C꜀) = 0.385 J/gºC

Heat lost by copper = heat gained by water

Divide both side by 95.48

Collect like terms

T꜀ = 335.6 °C

Therefore, the initial temperature of the piece of copper is 335.6 °C.

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335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

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

Given:

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

Find: Tc

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

Tc = 335

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.