How much heat is required to raise the temperature of 10.3 grams of water from 41.3°C to 97.2°C?

0 ⤊

How much heat is required to raise the temperature of 10.3 grams of water from 41.3°C to 97.2°C?

The specific heat of water is 1.00 cal/g-°C.

2007-10-05 03:32:23 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

3 answers

Q= m.s.dT where Q is the quantity of heat, m the mass of water, s the specific heat and dT the difference in temperature.

dT = 97.2 - 41.3 = 45.9

m = 10.3

s = 1

Q = 472.77 calories

2007-10-05 03:39:09 · answer #1 · answered by Swamy 7 · 0⤊ 0⤋

msdt is the formula for the warmth required to advance the temperature of 400grams of lead from 5 levels to a hundred sixty five levels. so 4 hundred*one hundred sixty*0.03= 1920calories. in case you desire answer in joules, multiply it with 4.2

2016-12-14 08:15:38 · answer #2 · answered by ? 4 · 0⤊ 0⤋

10.3X4.2X (97.2 -41.3) J. CACULATE

2007-10-05 03:40:43 · answer #3 · answered by Apparao V 4 · 0⤊ 0⤋