Yes, your approach and calculations look correct. Good job!

[STEMucator]

Homework Statement

How much heat is required to heat 0.2kg of ice from -20°C to 30°C?

Homework Equations

Total heat lost or gained : $Q = mcΔt$ where $Δt$ is the change in temperature and c is the heat capacity.

Latent Heat Fusion : $Q = mL_f$

The Attempt at a Solution

From my understanding, I think I need to break this problem down into states.

The ice will start melting at 0°C ( since ice is just condensed water ). So the ice must be heated from -20°C to 0°C for it to melt.

Then we must use Latent Heat to melt the ice.

Finally we would have to heat the water from 0°C to 30°C.

So the amount of heat we would need can be expressed by the equation :

$Q_T = Q_{Ice} + Q_{Melt} + Q_{Water} = mc_{Ice}Δt_{Ice} + mL_f + mc_{Water}Δt_{Water}$

Then plugging and chugging I get :

$(0.2 kg)(2100 J/kg°C)(0°C + 20°C) + (0.2 kg)(330000 J/kg) + (0.2)(4200 J/kg°C)(30°C - 0°C) = 99600J$

Therefore 99600J of heat would be required to heat the ice from its original state to a heated state of water.

Does this look okay?

 

[TSny]

Looks good!

 

[Doc Al]

Yep, looks good to me as well.

 

[BruceW]

yep. looks good to me too. Although you have used 2 significant figures in your values, so does it make sense to have 3 significant figures in your answer?

edit: hehe, everyone seems to have posted a reply near-simultaneously.

 

[STEMucator]

edit: hehe, everyone seems to have posted a reply near-simultaneously.

You're all so fast lol. Thanks for the confirm guys :).

 

[barryj]

"edit: hehe, everyone seems to have posted a reply near-simultaneously. " It's because it was an easy problem :-)