A pot on the stove contains 200 g of water at 20°C. An unknown mass ofice that i

Question

A pot on the stove contains 200 g of water at 20°C. An unknown mass ofice that is originally at-10°C is placed in an identical pot on the stove. Heat is then added to the two samples of water at precisely the same constant rate. Assume that this heat is transferred immediately to the ice or water (in other words, neglect the increase in temperature for the pot). We will also neglect evaporation. The ice melts and becomes water, and you observe that both samples of water reach 65.0°C at the same time. (a) (1 point) How does the mass of the original piece of ice in the second pot compare to the mass of the water in the first pot? Select one answer [ ] The mass of the ice is smaller than the mass of the water. [ ] The mass of the ice is the same as the mass of the water. The mass of the ice is larger than the mass of the water. [ ] There's not enough information to answer this question (b) (1 point) Which system will reach 80°C first? Select one answer. The first pot, which had the 20°C water in it at the beginning. [ ] The second pot, which had the ice in it at the beginning [ ] When the pots reach 65.0°C, they both contain water, so they will both reach 80°C at the same time. [ ] There's not enough information to answer this question (c) (1.5 points) Solve for the mass of the ice that was originally in the second pot. Give the answer in grams Useful info: the specific heat of liquid water is 4186 J/(kg °C), and of solid water is 2060 J/(kg °C). The latent heat of fusion ofwater is 3.35 x 105J/kg

Explanation / Answer

given
m_water = 200 g = 0.2 kg T1 = 20 C

T2 = -10 C

a) The mass of the ice is smaller than the mass of the water

b) The second pot, which had the ice in it at the begining.

c)

heat gained by ice in the second pot = heat gained by water in first pot

m_Ice*C_Ice*(0 - (-10)) + m_Ice*Lf + m_Ice*C_water*(65 - 0) = m_water*C_water*(65 - 20)

m_Ice*2060*10 + m_Ice*3.35*10^5 + m_Ice*4186*65 = 0.2*4186*45

m_Ice*(2060*10 + 3.35*10^5 + 4186*65) = 0.2*4186*45

m_Ice = 0.2*4186*45/(2060*10 + 3.35*10^5 + 4186*65)

= 0.0600 kg

= 60.0 grams <<<<<<<<<-----------------Answer

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