Use the heat equation to calculate the energy or calories lost when (8.8x10^1) g of water cools from 45 °C to 25 °C. Report answer to correct sig figs and in scientific notation.

Question 17 of 26 What is the heat needed to heat 150 g of water from 10.0°C to 100°C, (specific heat of water = 4186 J/kg-°C)? Select the correct response: 68.2 56.5 44.6 75.1 37.2 ㅋ

the laboratory a student finds that it takes 101 Joules to increase the temperature of 13.6 grams of solid iodine from 20.6 to 39.3 degrees Celsius. e specific heat of iodine calculated from her data is J/g°C.

Learning Goal: To understand the concepts of heat capacity, specific heat, and molar heat capacity. Heat capacity, C, is the amount of energy required to raise the temperature of a substance by exactly one degree Celsius. The energy needed to warm an object increases as the mass of that object increases. We see this in our everyday life. For example, we know that it takes much more energy to heat a large tank of water than a small cup. Because of this dependence on mass, experimentally determined heat capacities are always reported in terms of the amount of the substance that is heated. One method is to report how much energy it takes to raise the temperature of one mole of a substance by exactly one degree Celsius. This value is the molar heat capacity, which has the symbol Cp. The molar heat capacity is given in the units J mol- °C-1.A second method is to report how much energy it takes to raise the temperature of one gram of a substance by exactly one degree Celsius. This value is…