Question

A 50 g ice cube floats in 195 g of water in a 100 g copper cup; all are at a temperature of 0°C. A piece of lead at 96°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead?

Answer 1

Answer:

The mass of the lead will be "1.127 kg".

Explanation:

The given values are:

(Ice) m₁ = 50 g i.e.,

0.050 kg

(Water) m₂ = 195 g i.e.,

0.190 kg

(Copper cup) m₃ = 100 g i.e.,

0.100 kg

m₁, m₂ and m₃ at temperature,

t₁ = 0°C

Temperature of lead,

t₂ = 96°C

Temperature of Final equilibrium,

t₃ = 12°C

Let m₄ be the mass of the lead.

On applying formula, we get

⇒  $m_{1}L+m_{1}s_{1} \Delta t+m_{2}s_{2} \Delta t+m_{2}s_{2} \Delta t=m_{4}s_{4} \Delta t$

On putting the estimated values, we get

⇒  $(0.050)(334)+(0.050)(4186)(12-0)+(0.190)(4186)(12-0)+(0.100)(387)(12-0)=m_{4} (128)(96-12)$

⇒  $16.7+2511.6+9544.08+50.7=10752\times m_{4}$

⇒  $12,123.08=10752\times m_{4}$

⇒  $m_{4}=\frac{12,123.08}{10752}$

⇒  $m_{4}=1.127 \ kg$