A 450g sample of liquid water at 250 C is heated by the addi

A 4.50-g sample of liquid water at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the water is ________ °C. The specific heat capacity of liquid water is 4.18 J/g-K.

Solution

Answer:-

Therefore the formula is,

                                                     q = m × C × T

                                        where,

q = the heat absorbed / given off

                                       m = the mass of the sample

c = the specific heat of the substance

                                    T = the change in temperature

                                              Final temperature – initial temperature

Plug all the value you have in above equation,

145J = 4.50 g × 4.18 J/g × (T2 – 25)

145J = 18.81(T2 – 25)

145J = 18.81T2 -470.25

145 + 470.25 = 18.81T2

615.25 = 18.81T2

T2 = 615.25 / 18.81

T2 = 32.7