Answer:
The final temperature is 47.79 °C
Explanation:
Step 1: Data given
Sample 1 has a volume of 290.0 mL
Temperature of sample 1 = 25.00 °C
Sample 2 has a volume of 140.00 mL
Temperature of sample 2 = 95.00 °C
Step 2: Calculate the final temperature
Heat lost = heat gained
Qlost = -Qgained
Q = m*c* ΔT
Q(sample1) = -Q(sample2)
m(sample1) * c(sample1) * ΔT(sample1) = -m(sample2)*c(sample2) *ΔT(sample2)
⇒with m(sample1) = the mass of sample 1 = 290.0 mL * 1g/mL = 290 grams
⇒with c(sample 1) = the specific heat of water = c(sample 2)
⇒with ΔT(sample 1) = the change of temperature = T2 - T1 = T2 - 25.00 °C
⇒with m(sample2) = the mass of sample 2 = 140.0 mL * 1g/mL = 140 grams
⇒with c(sample2) = the specific heat of water = c(sample1)
⇒with ΔT(sample2) = T2 -T1 = T2 - 95.00°C
m(sample1) * ΔT(sample1) = -m(sample2)*ΔT(sample2)
290 *(T2-25.0) = -140 *(T2 - 95.0)
290 T2 - 7250 = -140 T2 + 13300
430 T2 = 20550
T2 = 47.79 °C
The final temperature is 47.79 °C
