Question

If 1,079.75 Joules of heat are added to 77.75 grams of water, by how many degrees Celsius would the water increase? Assume water is in the liquid state.

Answer 1

If 1,079.75 Joules of heat are added to 77.75 grams of water, by 3.32 degrees Celsius the temperature of water will increase

`\Delta T=3.32 C^(0)`

Step-by-step explanation:

`q = mC\Delta T`

Here , q = heat added / removed from the substance

m = mass of the substance taken

`\Delta T` = Change in temperature

C = specific heat capacity of the substance

In liquid state the value of C for water is :

`4.18 J/gC^(0)`

Given values :

q = 1079.75 J

m = 77.75 gram

Insert the value of C, m , q in the given equation

`q = mC\Delta T`

on transposing ,

`\Delta T=(q)/(mC)`

`\Delta T=(1079.75)/(77.75* 4.18)`

`\Delta T=3.32 C^(0)`