Heat lost by lead= heat gained by water
=> m_{lead} x c_{lead} x (T_{f} -T_{i}) = m_{water} x C_{water} x ( T_{f} - T_{i})
=> 34.1 g x 0.160 J/g^{0}C x( 96.9 -27.6)^{0}C = m_{water} x 4.184 J/g^{0}C x ( 27.6 - 21.1)^{0}C
=> 378.1008 J = m_{water} x 27.196 J/g
=> m_{water} =( 378.1008 /27.196) g
=> m_{water} = 13.9 g = 14 g ( 2 sig. figures)
Guestion 4 (1 point) A 34.1 gram piece of lead at 96.9。C is placed in a...
Question 21 (1 point) ✓ Saved A 40.1 gram piece of lead at 97.7 °C is placed in a calorimeter containing water at 21.7 °C. If the temperature at equilibrium is 27.6°C, what is the mass of the water? Report your answer in grams to one decimal place. Your Answer: 8 Answer units
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A piece of metal with a specific heat of 1.29 J/g°C is heated to 126.6°C and then placed in 133.868 g of water which is at a temperature of 10.9 °C. After a minute, the temperature of the water has stopped changing and is now 45.6°C. Assuming that there are no heat losses to the container or surroundings, what is the mass of the piece of metal in grams? Assume that water has a specific heat of 4.184 J/g°C. Enter...
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4. A 0.500 kg piece of copper at an initial temperature of 20.0°C is placed in a water bath and the temperature of the metal is raised to 100.0°C. Note: The specific heat capacity of copper is 385J/kg K and the latent heat of fusion is 2.07x105J/kg. a. How much heat was required to raise the temperature of the copper? e. An identical piece of heated copper (at 100.0°C) is placed in a calorimeter containing 0.500 kg of an unknown...
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