How much heat needs to be spent to bring 7 kg water taken at a temperature of 0 degrees

How much heat needs to be spent to bring 7 kg water taken at a temperature of 0 degrees to a boil and then completely evaporate it?

m = 7 kg. t1 = 0 “C. t2 = 100” C. C = 4200 J / kg * “C. λ = 2.3 * 10 ^ 6 J / kg. Q -? The amount of heat Q for complete evaporation of water will be the sum: Q = Q1 + Q2, where Q1 is the amount of heat for heating water from t1 = 0 “C to t2 = 100” C, Q2 is the amount of heat for evaporation at the boiling point. Q1 = C * m * (t2 – t1). Q1 = 4200 J / kg * “C * 7 kg * (100 “C – 0” C) = 2940000 J. Q2 = λ * m. Q2 = 2.3 * 10 ^ 6 J / kg * 7 kg = 16100000 J. Q = 2940000 J + 16100000 J = 19040000 J = 1904 * 10 ^ 4 J.
Answer: to evaporate water, you need Q = 1904 * 10 ^ 4 J of thermal energy.




One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.

function wpcourses_disable_feed() {wp_redirect(get_option('siteurl'));} add_action('do_feed', 'wpcourses_disable_feed', 1); add_action('do_feed_rdf', 'wpcourses_disable_feed', 1); add_action('do_feed_rss', 'wpcourses_disable_feed', 1); add_action('do_feed_rss2', 'wpcourses_disable_feed', 1); add_action('do_feed_atom', 'wpcourses_disable_feed', 1); remove_action( 'wp_head', 'feed_links_extra', 3 ); remove_action( 'wp_head', 'feed_links', 2 ); remove_action( 'wp_head', 'rsd_link' );