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kondor19780726
3 months 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 [3.4K]3 months 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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