Question

If 0.959J of heat is added to 0.88g of water, how much will its temperature increase

Answer 1

0.260 Celsius

Step-by-step explanation:

q =c x m x (T2-T1)

c - specific heat of water 4.186 J/g.C

T2-T1 = q /(c x m) = 0.959 /(4.186 x 0.88) = 0.959/3.68 =0.260 C