A T-bone steak provides 6.90 x 10² food Calories (Cal). A food Calorie is equivalent to 4180 J of heat energy. The heat energy provided by the steak would be sufficient to heat 59.1 kg of water how many Celsius degrees. The specific heat of water is 4.18 J/g °C.

Question

Quizplus · Accepted Answer

The heat energy from the steak is 6.90 x 10^2 Cal x 4180 J/Cal = 2.8842 x 10^6 J. Using the formula q = mcΔT, where q is the heat energy, m is the mass of water (59.1 kg = 59100 g), and c is the specific heat capacity of water (4.18 J/g°C), we solve for ΔT: ΔT = q / (mc) = 2.8842 x 10^6 J / (59100 g x 4.18 J/g°C) = 11.7 °C.

A T-Bone Steak Provides 6