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Imagine that your water heater has broken, but you want to take a bath. You fill

ID: 512163 • Letter: I

Question

Imagine that your water heater has broken, but you want to take a bath. You fill your bathtub with 25 kg of room-temperature water (about 25 C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.

Part A How much boiling water would you need to raise the bath to body temperature (about 37 C)? Assume that no heat is transferred to the surrounding environment. Express your answer to two significant figures and include the appropriate units. =4.80 kg (please show me equation(s) used to get this answer please)

Part B The amount of boiling water required to raise the temperature of 25.0 kg of water in the bath to body temperature is 4.80 kg. In this process, the heat lost by the boiling water is equal to the heat gained by the room-temperature water. How much heat was transferred in this process? Express your answer to four significant figures and include the appropriate units.

Explanation / Answer

Ans. Let the amount of boiling water (1000C) required = X gram

At equilibrium temperature (370C) heat gained by 25.0 L water at 250C is equal to heat lost by X kg boiling water.

The density of water is assumed to be 1.000 g/ mL or 1.000 kg/ L.

q = m x s x dT                      - equation 1   

Where,

m = mass of water in gram

s = specific heat of water = 4.184 J g-10C-1

dT = change in temperature = T1- T1

Heat gained by 25.0 L water at 250C = Heat lost by X kg boiling water

Or, q1 (25.0 L water at 250C) = q2 (X kg boiling water)

Or, 25000 g x (4.184 J g-10C-1) x (370C – 250C) = X x (4.184 J g-10C-1) x (1000C – 370C)

Or, 25000 g x 120C = X x 630C

Or, X = (25000 g x 120C)/ 630C = 4761.90 g

Thus, required mass of boiling water = 4761.90 g = 4.761 kg = 4.8 kg (approx.)

Part B. Amount of heat transferred = Amount of heat gained by cold water = Amount of heat lost be hot water.

Or, Amount of heat transferred = Amount of heat gained by cold water

                                                = 25000 g x (4.184 J g-10C-1) x 120C

                                                = 1255200 J

                                                = 1255.200 kJ                                   ; [1 kJ = 1000 J]

                                                = 1255 kJ