Answer:
The specific heat of the metal is 0.485 J/g°C
Explanation:
<u>Step 1:</u> Data given
Mass of the piece of metal = 80.0 grams
Mass of the water = 125 grams
Initial temperature of the metal = 88.0 °C
Initial temperature of water =20.0 °C
Final temperature = 24.7 °C
pecific heat of water is 4.18 J/g*°C
<u>Step 2:</u> Calculate specific heat of the metal
Qgained = -Qlost
Q =m*c*ΔT
Qwater = - Qmetal
m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)
with mass of water = 125 grams
with c( water) = 4.18 J/g°C
with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C
with mass of metal = 80.0 grams
with c(metal) = TO BE DETERMINED
with ΔT(metal) = 24.7 - 88.0 = -63.3 °C
125*4.18*4.7 = -80 * C(metal) * -63.3
2455.75 = -80 * C(metal) * -63.3
C(metal) = 2455.75 / (-80*-63.3)
C(metal) = 0.485 J/g°C
The specific heat of the metal is 0.485 J/g°C
