How many grams of ice at 0 will be melted in bringing down a 59 fever from 40C to 39C?

To calculate the amount of ice required, we first need to determine the heat that needs to be removed from the body. We can use the formula:

$$Q = mC\Delta T$$

where,

$$Q$$ is the heat required (in Joules)

$m$ is the mass of the substance (in kilograms)

C is the specific heat capacity of the substance (in Joules per kilogram per degree Celsius)

$$\Delta T$$ is the change in temperature (in degrees Celsius)

In this case, the mass of the body is not given, so we will assume an average mass of 70 kg. The specific heat capacity of the human body is approximately 3.47 kJ/kg/°C. The change in temperature is 40°C - 39°C = 1°C. Plugging these values into the formula, we get:

$$Q = (70 kg)(3.47 kJ/kg/°C)(1°C) = 242.9 kJ$$

Next, we need to determine how much ice is required to absorb this heat. The heat of fusion of ice is 334 kJ/kg. This means that it takes 334 kJ of heat to melt 1 kg of ice at 0°C. Therefore, the amount of ice required is:

$$m = \frac{Q}{L_f} = \frac{242.9 kJ}{334 kJ/kg} = 0.727 kg$$

Therefore, 0.727 kg or 727 grams of ice at 0°C will be required to bring down a 59 fever from 40°C to 39°C.