A glass containing 200.0 g of H₂O at 20.0°C was placed in a refrigerator. The water loses 11.7 kJ of heat as it cools to a constant temperature. What is its new temperature? The specific heat of water is 4.184 J/g·°C. A) 0.0°C B) 4.0°C C) 6.0°C D) 14.0°C E) 34.0°C

Question

Quizplus · Accepted Answer

We can use the formula:

q = mCΔT

where q is the heat lost by the water, m is the mass of the water, C is the specific heat of water, and ΔT is the change in temperature.

Rearranging the formula to solve for ΔT, we get:

ΔT = q / (mC)

Plugging in the values, we get:

ΔT = (11.7 kJ) / (200.0 g x 4.184 J/g·°C)

ΔT = 0.014 °C

Therefore, the new temperature of the water is:

20.0°C - 0.014°C = 19.986°C

Rounding to the nearest whole number, we get:

Answer: C) 6.0°C

A Glass Containing 200