Question

Calculate the heat flow when 18.7 g of water is heated from 25 degrees celcius to 68.2 degrees celcius (specific heat capacity for water=4.184J/g.k

Answer 1

Answer – We are given, mass of water = 18.7 g , initial temp, ti = 25^oC , Final temp, Tf = 68.2^oC

Specific heat capacity = 4.184 J/g^oC.

We know formula for calculating the heat

Heat, q = m x C x ∆T

Where, q – heat lost or absorbed

C- Specific heat capacity

∆T – Change in temperature

Now plugging the values in the formula

q = 18.7 g x 4.184 J/g^oC x (68.2 – 25)^oC

    = 3380 J

So, 3380 J heat flow when 18.7 g of water is heated from 25 degrees celcius to 68.2 degrees celcius.