Question

What is the final temperature of 25.0 grams of water at 22°c after it absorbs 454 j of heat? remember: the specific heat of water is 4.184 j/g∙°c\

Answer 1

Q = K M (T2 - T1) where Q is the heat absorbed

T2 = Q / (K * M) + T1

T2 = 454 / (4.184 * 25) + 22 = 4.34 + 22 = 26.34 deg