how many joules are needed to warm 25.5 grams of water from 14.0°C to 22.5°C? the heat capacity of liquid water is 4.184J/g°C

Q = 906.882 J

Explanation:

Given data:

Mass of water = 25.5 g

Initial temperature = 14.0 °C

Final temperature = 22.5 °C

Specific heat capacity of water = 4.184 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 22.5 °C - 14.0 °C

ΔT = 8.5 °C

Q = 25.5 g × 4.184 J/g.°C ×  8.5 °C

Q = 906.882 J

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.