The temperature of the water drops from 93.0°C to 78.0°C. What quantity of heat energy did the piece of aluminum absorb?The specific heat of Aluminum is 0.215 cal/g-°C. The calories absorbed (q) would be q = m * c * (T2 -T1) = 22.0 g * 0.215 cal/g-°C * (78.0 °C - 0.3 °C) If you have a different value for the Specific heat of Aluminum, use it in the formula for your calculations. If you knew the mass of the water, you could check your calculations because the heat gained by the Al would be equal to the heat lost by the water, using q = m * c * (T2 -T1), where m is the mass of the water, c = 1.00 calorie/gram °C, T2 = 78.0 °C and T1 = 93.0 °C. The sign of q will be negative.Other related question