Heat (q)= mc(∆T)
= 110×1.0×(79-15)
=7.0 kilocalories
So answer is
7.0 kilocalories
II ReviewI ConstantsPeriodic Table Use the heat equation to calculate the energy for each of the...
3.37 Use the heat equation to calculate the energy for each of the following (see Table 3.11): a. calories to heat 8.5 g of water from 15 °C to 36 °C b. 2600 joules lost when 25 g of water cools from 86°C to 61 °C c. 9.3 kilocalories to heat 150 g of water from 15 °C to 77 °C d. kilojoules to heat 175 g of copper from 28 °C to 188 °C 3.38 Use the beat. cal/g...
Use the heat equation to calculate the energy, in joules and calories, for each of the following (see the table): Specific Heats for Some Substances Substance cal/g ∘Ccal/g ∘C J/g ∘CJ/g ∘C Elements Aluminum, Al(s)Al(s) 0.214 0.897 Copper, Cu(s)Cu(s) 0.0920 0.385 Gold, Au(s)Au(s) 0.0308 0.129 Iron, Fe(s)Fe(s) 0.108 0.452 Silver, Ag(s)Ag(s) 0.0562 0.235 Titanium, Ti(s)Ti(s) 0.125 0.523 Compounds Ammonia, NH3(s)NH3(s) 0.488 2.04 Ethanol, C2H6O(s)C2H6O(s) 0.588 2.46 Sodium chloride, NaCl(s)NaCl(s) 0.207 0.864 Water, H2O(s)H2O(s) 1.00 4.184 Water, H2O(s)H2O(s) 0.485 2.03 a.)...
A Revi Use molar volume to calculate each of the following at STP: Part A the volume, in liters, occupied by 3.50 moles of N2 gas 190 AS O O ? V = Submit Request Answer Part B the number of moles of CO2 in 2.10 L of CO2 gas Co Agon O O ? V = mol Submit Request Answer Problem 3.38 Review I Constants | Part A calories lost when 95 g of water cools from 45°C to...