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

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

Determination of the mass of water

Volume of water = 390 mL

Density of water = 1 g/mL

Cross multiply

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

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

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

spectral lines lyman is one set

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2