A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.55 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).

1.0188 J/g*C

Explanation:

Using the formula; Q = m × c × ∆T

Q(water) = -Q(metal)

m × c × ∆T (water) = -{m × c × ∆T (metal)}

According to this question,

mass of metal = 25g

initial temp of metal = 99°C

mass of water = 50g

initial temp of water = 10.55°C

final temperature of the system = 20.15°C

c of water = 4.184 J/g*C

50 × 4.184 × (20.15 - 10.55) = 25 × c × (20.15 - 99)

209.2 × 9.6 = 25c × -78.85

2008.32 = -1971.25c

c = 2008.32 ÷ 1971.25

c of metal = 1.0188 J/g*C

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'