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

each chemical family is represented as a group or column on the periodic table. based on chemical properties, there are 6 types of chemical families. group 1 represents alkali metals, group 2 represents alkaline earth metals, groups 3 to 12 represent transition metals, groups 13 to 16 represent non-metals and other metals, group 17 represents halogens and group 18 represents noble gases.