Calculating Heat Required for Water Temperature Change

[kimkibun]

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:
m_Al=178g
m_H2O=90g
T₁=15.5°C
T₂=85°C
c_H2O=1cal/g°C

Homework Equations

Im planning to use this formula for heat quantity

Q_H2O=m_H2Oc_H2OΔT_H2O

where ΔT_H2O=T₂-T₁

The Attempt at a Solution

Since what we need is Q_H2O i just ignore m_Al, using the above formula i got

Q_H2O=(90g)(1cal/g°C)(85-15.5)°C=6300cal

is this correct?

 

[gneill]

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:
m_Al=178g
m_H2O=90g
T₁=15.5°C
T₂=85°C
c_H2O=1cal/g°C

Homework Equations

Im planning to use this formula for heat quantity

Q_H2O=m_H2Oc_H2OΔT_H2O

where ΔT_H2O=T₂-T₁

The Attempt at a Solution

Since what we need is Q_H2O i just ignore m_Al, using the above formula i got

Q_H2O=(90g)(1cal/g°C)(85-15.5)°C=6300cal

is this correct?

You're halfway there What about the enclosing aluminum vessel? Doesn't its temperature have to be raised to the final value also?

 

[kimkibun]

You're halfway there What about the enclosing aluminum vessel? Doesn't its temperature have to be raised to the final value also?

do i need to get Q_Al?

 

[gneill]

do i need to get Q_Al?

I don't see how you can heat the water without also heating the vessel that contains it. So I'd have to say yes, you should add the heat required to raise the temperature of the aluminum vessel too.