Calculate the heat capacity of 1,343 g of lead, given that 45 j is needed to raise the temperature by 29. 8 ∘c.

The specific heat capacity of 1,343 grams of lead, given that 45 joules is needed to raise the temperature by 29.8°C, is equal to 0.00112 J/(g · °C).What is specific heat capacity?Before you have an understanding of what specific heat capacity is, you have to know the concept of calorimetry first. Calorimetry is a measurement of different changes in the state variables of a body for deriving the heat transfer related to the changes of its states. Mathematically speaking, we can define calorimetery as Q = mcΔT where Q represents heat energy, m represents mass, c represents specific heat capacity, and ΔT represents change in temperature. With a little bit of rearrangement, the formula of calorimetry can be used to find the specific heat energy:[tex]\frac{Q}{m\Delta T}=\frac{mc\Delta T}{m\Delta T}\\\\\frac{Q}{m\Delta T}=c\\\\c=\frac{Q}{m\Delta T}[/tex]The specific heat energy can then be calculated as follows (suppose that Q = 45 joules, m = 1,343 grams and Δt = 29.8°C):[tex]c=\frac{Q}{m\Delta T}\\\\c=\frac{45}{1,343\times 29.8}\\\\c\approx 0.00112[/tex]The specific heat energy is confirmed to be approximately equal to 0.00112 J/(g · °C).Learn more about calorimetry here: https://brainly.com/question/8168263#SPJ4