Question

Problem 2 (20 pts): You get home one day to find that your roommate is ill and has a dangerously high temperature of 106 F # 41.0 C. In order to quickly reduce his temperature, you put him in the bath tub with 8.00 L of water at 25.0C.

a) What is the mass of the water in the tub in units of kg? 

b) What is the mass of ice that you would need to add to the tub in order for your roommate and the water to reach an equilibrium temperature of 98.6F 37.0 C, which is normal body temperature, once the ice has entirely melted? Calculate the mass of ice in kg. Assume the ice has an initial temperature of 0.00 C and your roommate has a mass of 70.0 kg. For reference, here are the values of some relevant specific heats:

Answer 1

a) density of water at 25^oC is 997 kg/m³

given volume, = 8 L = 0.008 m³ {since 1 L =10^-3 m³}

we know, mass/volume =density

massof water, m_w = 0.008*997 =7.976 kg

b) we will use conservation of energy (heat )

heat lost by the body= heat gained by the water and the ice

so, heat lost by body = m_bc_b*dT_b = 70*3470*(41-37) =971600 J {dT is the change in temperature}

heat lost by water = m_wc_w*dT_w = 7.976*4186*(37-25) =400650.432 J

heat lost by ice is in two stages, first is through phase transformation(latent heat of fusion), and then sensible heat

=m* $\lambda$ + m*c_i*dT_i = m*334 + m*4186*(37-0) =155216*m  J

putting all these values in the energy balance

971600 = 400650.432 + 155216*m

m= 3.678 kg

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