Solving for Melting Ice: 500g Lead & -10C Ice Block

[ally.hady]

Homework Statement

a 500 g lead mass is heated to 150 C and placed on a block of ice at -10 C. how much ice, if any will melt? 52 kg = block of ice

Homework Equations

The Attempt at a Solution

i didint know where to start so i did qmcdeltaT + qmcdeltaT = 0, and i failed at it i asked a friend and they said u have to ur heat of fusion some how because ice is melting please show me!

 

[gneill]

How much heat would have to be put into the 52 kg block of ice in order to bring its temperature up to zero C where melting can occur?

 

[ally.hady]

q=mcDeltaT

Q=52kg(2100 j/kg C) (0C-(-10C))
Q= 1092 Kj

thats the amount of heat needed to bring its temperature to 0 degree now where do i go from here?

 

[gneill]

Well, how about checking to see how much heat is available from the lead? The lead is starting at 150C. How much heat could it provide if it were to be brought all the way down to zero C?

 

[ally.hady]

but i don't have the specific heat capactiy on the sheet given to me so how do i find the heat?

 

[gneill]

but i don't have the specific heat capactiy on the sheet given to me so how do i find the heat?

Well, that is a problem. You could look it up (the easy way!).

Or, you could present an argument... The specific heats of all metals are less than that of water. The nearest contender is Lithium with a specific heat of 3.58 kJ/kg/K, with water being at 4.187 kJ/kg/K. Almost all the rest of the metals are less than 1 kJ/kg/K (and in particular, lead comes in at a measly 0.128 kJ/kg/K).

Given the mass of lead provided, estimate the maximum heat it might be able to provide to warm the ice (from -10° to the melting point) if it's heat capacity were comparable to that of water (definitely erring on the conservative side!). Is it enough to warm the block to melting point?

 

[ally.hady]

wow ur smart, ok i will do that