Answer:
The equilibrium temperature of water is 25.6 °C
Explanation:
Step 1: Data given
Mass of the sample of brass = 25.0 grams
The specific heat capacity = 0.375 J/g°C
Mass of water = 250.0 grams
Temperature of water = 25.0 °C
The initial temperature of the brass is 96.7°C
Step 2: Calculate the equilibrium temperature
Heat lost = heat gained
Q(sample) = -Q(water)
Q = m*C* ΔT
m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)
⇒m(sample) = the mass of the sample of brass = 25.0 grams
⇒with c(sample) =The specific heat capacity = 0.375 J/g°C
⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C
⇒with m(water) = the mass of the water = 250.0 grams
⇒with c(water) = the specific heat capacity = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C
25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)
9.375T2 - 906.56 = -1046T2 + 26150
9.375T2 + 1046T2 = 26150 + 906.56
1055.375T2 = 27056.26
T2 = 25.6 °C
The equilibrium temperature of water is 25.6 °C
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