How much heat is required to heat 200 g of alcohol from 18 degrees to 48 ° C in a glass flask weighing 50 g?

Initial data: m1 (mass of alcohol) = 200 g = 0.2 kg; t (temperature before heating) = 18 ºС; tк (final temperature) = 48 ºС; m2 (mass of glass bulb) = 50 g = 0.05 kg.

Reference data: C1 (specific heat capacity of alcohol) = 2500 J / (kg * K); C2 (specific heat capacity of glass) = 840 J / (kg * K).

Required amount of heat: Q = Q1 (heating alcohol) + Q2 (heating the flask) = C1 * m1 * (tk – t) + C2 * m2 * (tk – t) = (tk – t) * (C1 * m1 + C2 * m2).

Let’s calculate: Q = (48 – 18) * (2500 * 0.2 + 840 * 0.05) = 16 260 J.