Two heat/temperature problems

[demode]

1. How many celsius degrees does the temperature of a mass of 1.00 kg of water fall for each 83.8 kJ of heat it loses.
2. Not so sure about EQUATIONS but I have my own way of going at it.
3. 83.8 kJ / 4.19 J * (1.00 kg * 1000) = 0.2 degrees celsius



1. How much heat is required to raise the temperature of a mass of 10.0 kg of water a total of 20.0 celsius degrees.
2. Not so sure about EQUATIONS but I have my own way of going at it.
3. IF it takes 4.19 j of heat to raise the temperature of 1 gram of water by one degree celsius, how do we factor in the larger mass of 10,000 grams and the larger temperature change of 20 celsius degrees?

 

[Hootenanny]

Your solution to the first question is correct. For the second, you should consider an equation relating the specific heat capacity of water, its mass and the change in temperature to thermal energy. I'm sure you will have been given some formulae in your notes/text.

 

[m2003]

1. The amount of heat required to raise the temperature of a substance can be calculated using the specific heat capacity of the substance. In the case of water, the specific heat capacity is 4.19 J/g°C. Therefore, the amount of heat required to raise the temperature of 10.0 kg of water by 20.0°C can be calculated as follows:

Q = m * c * ΔT

Where Q is the amount of heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Substituting the values, we get:

Q = (10.0 kg * 1000 g/kg) * (4.19 J/g°C) * (20.0°C) = 838,000 J or 838 kJ

Therefore, 838 kJ of heat is required to raise the temperature of 10.0 kg of water by 20.0°C.

2. It is important to use equations and proper units when solving scientific problems to ensure accurate and consistent results. In this case, using the specific heat capacity equation allows us to factor in the mass and temperature change of the water in a precise manner.