kimkibun said:Homework Statement
An aluminum vessel hose mass is 178g contains 90g of water at 15.5°C. How many calories of heat are required to bring the water to a final temperature of 85°C? The specific heat of water is 1cal/g°C.
Given:
mAl=178g
mH2O=90g
T1=15.5°C
T2=85°C
cH2O=1cal/g°C
Homework Equations
Im planning to use this formula for heat quantity
QH2O=mH2OcH2OΔTH2O
The Attempt at a Solution
The heat required to heat water can be calculated using the formula Q = mCΔT, where Q is the heat required, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature (in this case, from 15.5°C to 85°C).
The specific heat capacity of water is 4.186 joules/gram·degree Celsius. This means that it takes 4.186 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
The mass of water can be determined by knowing the volume and density of water. The formula for density is D = m/V, where D is density, m is mass, and V is volume. Since the density of water is approximately 1 gram/mL, you can assume that the mass and volume are equal in this case.
Yes, this calculation can be used for any type of water, as long as you use the correct specific heat capacity for that type of water. Different types of water (such as saltwater or freshwater) have different specific heat capacities, so it is important to use the appropriate value in the calculation.
This calculation is fairly accurate, as long as all the variables (mass, specific heat capacity, and change in temperature) are measured accurately. However, it is important to note that this calculation does not take into account any external factors, such as heat loss to the environment, which may affect the actual amount of heat required to heat the water.