if 4.0 grams of water at 1*c absorbs 33 joules of heat, what will be the change in temperature of the water? (

Answer 1

we have Q = mc * (t2 - t1) << c here is Specific heat capacity >>
=> t2 - t1 = Q / (mc) = 33 / (0.004 * 4200) = 55/28
=> t2 = 55/28 + t1 = 83/28 = approx. 2.96*C
thus the temperature of water will rise to 2.96*C