. The specific heat of lead is 0.030 cal / g . ° C.300 grams of lead shot at 100 ° C is mixed with 100 grams of water at 70 ° C in an insulated container . The final temperature of the mixture is&nbsp;

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. The specific heat of lead is 0.030 cal / g . ° C.300 grams of lead shot at 100 ° C is mixed with 100 grams of water at 70 ° C in an insulated container . The final temperature of the mixture is&amp;nbsp;

Mr Solver · Accepted Answer

To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the lead (shot) will be equal to the heat gained by the water.The heat lost by the lead can be calculated using the formula:Qlead=mlead·clead·ΔTleadwhere mlead is the mass of the lead, clead is the specific heat of lead, and ΔTlead is the change in temperature of the lead.Similarly, the heat gained by the water can be calculated using the formula:Qwater=mwater·cwater·ΔTwaterwhere mwater is the mass of the water, cwater is the specific heat of water, and ΔTwater is the change in temperature of the water.Since the system is insulated, the heat lost by the lead is equal to the heat gained by the water. Therefore, we can set Qlead equal to Qwater and solve for the final temperature of the mixture.Let&#039;s plug in the given values:Mass of lead, mlead=300gramsSpecific heat of lead, clead=0.030cal/g&amp;nbsp;°CInitial temperature of lead, Tlead=100°CMass of water, mwater=100gramsSpecific heat of water, cwater=1cal/g&amp;nbsp;°CInitial temperature of water, Twater=70°CUsing the equation Qlead=Qwater, we can calculate the final temperature of the mixture.

. The specific heat of lead is 0.030

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