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A 130 g copper bowl contains 100 g of water, both at 20.0°C. A very hot 420 g copper cylinder is dropped into the water, causing the water to boil, with 8.63 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.

Respuesta :

Answer:

a) 4652 cal

b) 8000 cal

Explanation:

Amount of heat transferred

Q1 = mL(v)

Q1 = 8.63 * 539

Q1 = 4652 cal

Amount of heat transferred to the water

Q2 = mcΔT

Q2 = 100 * 1 * (100 - 20)

Q2 = 8000 cal

Q = Q1 + Q2

Q = 4652 + 8000

Q = 12652 cal

b)

Heat transferred to the copper bowl

Q(b) = m(b) * c(b) * ΔT

Q(b) = 0.13 * 0.0923 * (100 - 20)

Q(b) = 0.96 cal

c)

Original heat of the cylinder

Q(c) = Q + Q(b)

m(c) * c(c) * ΔT = Q + Q(b), making ΔT subject of the formula

ΔT = (Q + Q(b))/ (m(c) * c(c))

ΔT = (12652 + 0.96) / (0.42 * 1)

ΔT = 12652.96/0.42

ΔT = 30126.1

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