a typical woman gives off heat at rate of about 8000 calories per hour. how long would a woman have to stay in a bath of 60L (60,000 g) of 26 C water in order to raise the water temperature to 30 c ? presume that all the heat given off by the woman is transferred to the water , and that the water does not lose any heat to the air. show your calculations..

To determine how long it would take for a woman to raise the temperature of the water from 26°C to 30°C, you need to calculate the amount of heat energy the woman can transfer to the water and then determine how long it would take for that heat energy to raise the temperature of the water, assuming no heat loss.

Step 1: Calculate the amount of heat energy the woman can transfer to the water:

The heat energy (Q) transferred to the water is given by the equation:

Q = mcΔT

Where:
Q = heat energy transferred (in calories)
m = mass of water (in grams)
c = specific heat capacity of water (in calories/gram °C)
ΔT = change in temperature (in °C)

Given:
m = 60,000 grams
c = 1 calorie/gram °C (specific heat capacity of water)
ΔT = 30°C - 26°C = 4°C

Plugging in the values:

Q = 60,000 grams * 1 calorie/gram °C * 4°C
Q = 240,000 calories

Step 2: Calculate the time required to transfer the heat energy to the water:

The heat transfer equation is:

Q = Pt

Where:
Q = heat energy transferred (in calories)
P = power (heat rate) at which the woman gives off heat (in calories/hour)
t = time (in hours)

Given:
P = 8,000 calories/hour
Q = 240,000 calories

Plugging in the values:

240,000 calories = (8,000 calories/hour) * t

Divide both sides by 8,000 calories/hour:

t = 240,000 calories / 8,000 calories/hour
t = 30 hours

Therefore, it would take approximately 30 hours for the woman to raise the water temperature from 26°C to 30°C, assuming no heat loss.