It’s number 4! I don’t know how to solve it. If you can, thank you and please show the steps!! **Student Name:****Problems:**1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not?2. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?3. A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. - (a) What is the specific heat of the substance? - (b) If it is one of the substances found in Table 9.1, what is its likely identity?4. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.5. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal. - (a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. - (b) The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer.6. Dissolving 3.0 g of CaCl₂(s) in 15.0 g of water in a calorimeter (Figure 9.12) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic?

We know that specific heat is the amount of heat required to raise the temperature of 1 g of substance by 10 C and it can be expressed as q=c X m X (Tfinal-Tinitial) ................1 Given that the specific heat of water and coffee is same and also the density of water and coffee is same = 1.0 g/mL Volume of coffee = 180 mL The final temperature of the resulting mixture = Tf = 600C Initial temperature of water, Ti water = 230C Initial temperature of coffee, Ti coffee = 950C we know that, the density of coffee is 1.0 g/mL so the mass of coffee will be =mcoffee =180g When we mix water with coffee, the temperature of water increases from 230C to 600C whereas the temperature of coffee decreases from 950C to 600C from the above changes we can say that the the final temperature of both water and coffee is same which is = 600C Now let's assume that, the total heat lost by coffee is equal to the total heat gain by water qcoffee = -qwater..............2 By comparing equation 1 and 2, we get ccoffee X mcoffee X (Tfinal -Tinitial)coffee = -[cwater X mwater X (Tfinal -Tinitial)water] By substituting values in the above equation, we get c X (180g) X ( 600C - 950C) = -[c X mwater X (600C -230C)] mwater=-[180g X -350C]370C so the mass of water= 170.27g we know that density given by Density=MassVolume...........3 Given density of water = 1.0 g / mL and mass of water is 170 .27g so volume of water can be calculated from equation 3 as Volume = Mass X Density = 170 .27g X 1.0 g /mL = 170.27mL Hence the volume of water at 230C that should be mixed with 180 mL of coffee at 950C to make a resulting mixture of 600C is 170.27mL