Question

Homework Name: 1) How much heat, in joules, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? 2) A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance? 3) A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0°C contained in a calorimeter like that shown in Figure. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal?

Answer 1

1) Heat that must be added mass x Specific heat x AT = 75.00 * 0.449 5(535-2550€ - 7°C = 50849.25 J Heat, Qamcat soo me mass c = specific heat capacity Ati temperature Change = final a Initial temperature specific heat, C = @ maT 2110 J 44.79 * (89:6– 23.2) °C = 10.711 5/9°C] Heat given by metal = heat absorbed by wath = (m Cat) water = 1009 * 4.184 L x6 goc 42_*(246–22:0) 11087.84 J)

So, -1087.84 J = 70.1 Again heat given by metal = (m CAT) metal 80:0) °C or -1087184 J = 70.09xcx (55.4°c) or c = 1087.84 J 70.09% (-55.4°C) = 0.281 5/gºc Thus, specific heat of metal= (0.281 J/goc)