If 50 g of lead (of specific heat 0.11 kcal/kg ∙ °C)at 100°C is put into 75 g of water (of specific heat 1.0 kcal/kg ∙ °C)at 0°C.What is the final temperature of the mixture?
A)2.0°C
B)6.8°C
C)25°C
D)50°C

Question

Quizplus · Accepted Answer

Using the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature, we can find the amount of heat lost by the lead and gained by the water:
Q_lead = m_lead × c_lead × ΔT_lead = 50 g × 0.11 kcal/kg ∙ °C × (100°C - T_final)
Q_water = m_water × c_water × ΔT_water = 75 g × 1.0 kcal/kg ∙ °C × (T_final - 0°C)
Since the heat lost by the lead must equal the heat gained by the water, we have:
Q_lead = Q_water
50 g × 0.11 kcal/kg ∙ °C × (100°C - T_final) = 75 g × 1.0 kcal/kg ∙ °C × (T_final - 0°C)
Solving for T_final, we get:
T_final = 6.8°C
Therefore, the answer is B.

If 50 G of Lead (Of Specific Heat 0