A 82-g cube of ice at 0°C is dropped into 1.0 kg of water that was originally at 84°C?
A 82-g cube of ice at 0°C is dropped into 1.0 kg of water that was originally at 84°C. What is the final temperature of the water after the ice has melted?
°C?
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released heat = absorbed heat
Qr=Qa
m1(T1-Tf)=m2c2(Tf-T)+m2λ
Tf=(m1c1T1+m2c2T2-m2λ)/(m1c1+m2c2)
1=water c1=4186 J/(kg K)
2=ice c2=2040 J/(kg K) λ=3,3x01^5 K/kg
then
Tf=(1*4186*357 + 0,082*2040*273 - 0,082*3,3x10^5) / (1*4186+0,082*2040)
Tf=337 K = 64 C