What would the expected temperature change be (in f) if 0.5 gram sample of water released 50.1 j of heat energy? the specific heat of liquid water 4.184 j/g

ΔT =  75.11 °F

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass= 0.5 g

heat released = 50.1 j

specific heat of water=  4.184 j/g .°C

temperature change =ΔT= ?

Solution:

Q = m × c × ΔT

50.1 j = 0.5 g × 4.184 j/g × ΔT

50.1 j = 2.092 j.°C × ΔT

50.1 j/ 2.092 j.°C = ΔT

23.95°C = ΔT

conversion of °C to °F

(23.95 × 9/5 ) + 32 = 75.11 °F

ΔT =  75.11 °F

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'