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

Answer:

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

$\Delta T=3.32 C^{0}$

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 ,

$\Delta T=\frac{q}{mC}$

$\Delta T=\frac{1079.75}{77.75\times 4.18}$

$\Delta T=3.32 C^{0}$