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A piece of metal with a mass of 1.9 kg, and specific heat of 200J/kg°C, and an initial temperature of 100°C is dropped into an insulated jar that contains liquid with a mass of 3.9 kg, a specific heat of 1000J/kg°C, and initial temperature of 0°C. The piece of metal is removed after 7 s, at which time its temperature is 20°C.

Required:
Find the temperature of the liquid after the metal is removed. Neglect any effects of heat transfer to the air or to the insulated jar. Answer in units of ∘C.

Sagot :

Answer:

7.79 °C

Explanation:

Given that

Mass of the metal, m = 1.9 kg

Specific heat of the metal, c(m) = 200 J/kgC

Initial temperature, T(0) = 100 °

Mass of liquid, M = 3.9 kg

Specific heat of liquid, c(l) = 1000 J/kgC

Initial temperature, T = 0 °C

Time taken, t = 7 s

Final temperature = 20 °C

To solve the question, we use the formula the sum of the heats gained is zero.

massmetal * cmetal (20-100) + massliquid * cliquid (T.f-0)=0

On solving for T.f

1.9 * 200 * (20 - 100) + 3.9 * 1000 * (t.f - 0) = 0

380 * -80 + 3900 * t.f = 0

-30400 + 3900t.f = 0

3900t.f = 30400

t.f = 30400 / 3900

t.f = 7.79 °C

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