If you combine 290.0 mL 290.0 mL of water at 25.00 ∘ C 25.00 ∘C and 140.0 mL 140.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

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

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong