A 11.8-gram sample of zinc is heated to 98.15°C in a boiling water bath. The hot metal is then immersed in 25.35 grams of water that is initially at 21.82°C. At what temperature will the metal and the water reach thermal equilibrium? The specific heat of zinc is 0.388 J/g·°C. The specific heat of water is 4.184 J/g·°C.

Question

Physics

Emely Marshall

10 December, 02:00

A 11.8-gram sample of zinc is heated to 98.15°C in a boiling water bath. The hot metal is then immersed in 25.35 grams of water that is initially at 21.82°C. At what temperature will the metal and the water reach thermal equilibrium? The specific heat of zinc is 0.388 J/g·°C. The specific heat of water is 4.184 J/g·°C.

+5

Answers (1)

Know the Answer?

Landin · Answer 1

T = 24.98 °C

Explanation:

from equilibrium condition we have:

[m_1*s*T]_{water} = [m_2*s*T]_{zinc}

Where

m_1, m_2 is mass of water and zinc respectively

s denote specific heat

let final equilibrium temperature is T

25.35*4.184*[T-21.82] = 11.8*0.388*[98.15-T]

106.06T - 2314.32 = 449.36 - 4.57T

110.63 T = 2763.68

T = 24.98 °C