We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)
The specific heat of the metal is 0.44 J/ (°C × g)
Answer:
Volume of water at this temperature is 27.2 mL
Explanation:
We know that 
Here density of water is 0.992 g/mL
Here mass of water is 27.0 g
So 
= 
= 27.2 mL
I am sure, the answer is variant B.
Number 3 i think is <span>d.heat moves from an object of higher temperature to an object of lower temperature</span>
The answer is B) fills all the space in its container