Question 1 of 25 How many degrees will 340 J raise the temperature of 6.8 g of water? (Refer to table of constants for water.) A. 28^circ C B. 50^circ C C. 81^circ C D. 12^circ C

To solve this problem, we need to use the formula for calculating the change in temperature of a substance when a certain amount of heat energy is added or removed. The formula is:<br /><br />ΔT = Q / (m × c)<br /><br />Where:<br />- ΔT is the change in temperature (in degrees Celsius)<br />- Q is the amount of heat energy added or removed (in joules)<br />- m is the mass of the substance (in grams)<br />- c is the specific heat capacity of the substance (in joules per gram per degree Celsius)<br /><br />Given information:<br />- Q = 340 J<br />- m = 6.8 g<br />- c = 4.18 J/g°C (specific heat capacity of water)<br /><br />Plugging in the values, we get:<br /><br />ΔT = 340 J / (6.8 g × 4.18 J/g°C)<br />ΔT = 340 J / 28.424 J/°C<br />ΔT = 11.94°C<br /><br />Therefore, the correct answer is:<br />D. $12^{\circ }C$