What is the temperature increase of 3.0 kg of water when heated by an 800-W immersion heater for 10 min? (c_w = 4186 J/kg · °C) A) 56°C B) 38°C C) 29°C D) 14°C

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat energy transferred to the water using the formula 
Q = P*t, where P is the power of the immersion heater and t is the time it is in operation. 
Here, P = 800 W and t = 10 min = 600 s. So, 
Q = 800 * 600 = 480000 J

Next, we need to calculate the change in temperature of the water using the formula 
Q = m*c*ΔT, where m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. 
Here, m = 3.0 kg and c = 4186 J/kg · °C. Substituting these values in the formula, we get 
480000 = 3.0 * 4186 * ΔT
ΔT = 40.79°C

Therefore, the temperature of the water increases by 40.79°C. The closest option is B) 38°C.

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