Answer:
Mass, m = 27g
Explanation:
Given the following data;
Initial Temperature, T1 = 25°C
Final temperature, T2 = 50°C
Quantity of heat = 2825J
Specific heat capacity of water = 4.184
Heat capacity is given by the formula;
[tex] Q = mcdt[/tex]
Where,
Making mass, m the subject of formula, we have;
[tex] m = \frac {Q}{cdt} [/tex]
Change in temperature, dt = T2 - T1
Change in temperature, dt = 50-25 = 25°C
Substituting into the equation, we have;
[tex] m = \frac {2825}{4.184 * 25} [/tex]
[tex] m = \frac {2825}{104.6} [/tex]
Mass, m = 27g
Therefore, the mass of water that can be added is 27 grams.