When 108 g of water at a temperature of 22.5?

ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCΔT 11466.58 Joules = 65.1g x 4.18 J / g C x ΔT 11466.58/(65.1gx4.18)=ΔT ΔT=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).

Try b? sorry if i’m incorrect

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.