Thermodynamics help?

Question

A water heater can generate 33,000 kJ/h. How much water can it heat from 15°C to 51°C per hour?

A small immersion heater is rated at 325 W. Estimate how long it will take to heat a cup of soup (assume this is 250 ml of water) from 20°C to 50°C?

Please try to show work too. Thanks

Answer 1

In both problems you must state the assumption that no heat is lost to the ambient, but all heat is applied to the water.

1)  The specific heat of water is 1 calorie/(gram °C) = 4.186 Joule/(gram °C)  (you really should memorize that one)

delta T = 36 °C

{[4.186 J / (X grams °C) * 36°C] / 33,000,000 J} * 36 = 1/Xgrams

X = 220 kg (per hour)

2)  325 Watts = 325 Joules/sec

250 ml of water = 250 g (at the temperatures given)

delta T = 30°C

4.186J /°Cg * 30°C * 250g / 325J/s = 96.6 seconds

.

Answer 2

Use the basic relation Q = m * Cp * dT

Q= 33000

Cp = Specific heat for water (4.186 KJ/kg C)

m = mass of water = ?

dT = Temperature difference = 51 - 15 = 36

33000 = m * 4.186 * 36

Solve for m

m = 218.99 kg of water can be heated per hour

2)  You need the specific heat for the soup. If it's not given you can assume the Cp for water (given above 4.186 KJ/kg C). Since volume of water is given, knowing the specific gravity you convert it to kg (0.25 kg). You can use the above formula, and then you will get the Q in KJ. Then you multiply that with 1000. You get J. Now Watt = J/s. Divide by 325.

Q = 0.25 * 4.186 * 30 = 31.395 KJ = 31395 J

Total heat required to up the soup temperature by 30 C is 31395 J. Now the heater is 325 J/s. So divide by 325

31395 / 325 = 96.6 seconds

Time required to heat the cup of soup is 96.6 seconds.