A10.0-gram sample of liquid water at 23.0°c absorbs 209 joules of heat.
what is the final temperature of the liquid water sample?

a) 5.0°c b) 28.0°c c) 18.0°c d) 50.0°c

The final temperature of the water is 28.0 °C

Explanation:

Step 1: Data given

Mass of liquid water = 10.0 grams

Temperature = 23.0 °C

Heat absorbed = 209 Joules

Since heat was absorbed by the water, you must have a positive value for  

Δ T

Step 2: Calculate final temperature

q = m*c* ΔT

⇒ with m = the mass of the water = 10.0 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = T2 -  T1 = T2 - 23.0 °C

⇒ with q = the heat absorbed = 209 Joule

209 = 10.0 * 4.184 * ΔT

ΔT = 5

ΔT = 5 = T2 - 23

T2 = 28 °C

The final temperature of the water is 28.0 °C

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml