Answer:
The specific heat of the alloy is 2.324 J/g°C
Explanation:
<u>Step 1:</u> Data given
Mass of water = 0.3 kg = 300 grams
Temperature of water = 20°C
Mass of alloy = 0.090 kg
Initial temperature of alloy = 55 °C
The final temperature = 25°C
The specific heat of water = 4.184 J/g°C
<u>Step 2:</u> Calculate the specific heat of alloy
Qlost = -Qwater
Qmetal = -Qwater
Q = m*c*ΔT
m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)
⇒ mass of alloy = 90 grams
⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED
⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C
⇒ mass of water = 300 grams
⇒ c(water) = the specific heat of water = 4.184 J/g°C
⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C
90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C
c(alloy) = 2.324 J/g°C
The specific heat of the alloy is 2.324 J/g°C
