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b) A student heats up a 5000 mathrm(~g) block of cion from 18^circ mathrm(C) to 46^circ mathrm(C) , transforming 62,8 mathrm(~kJ) of energy. Calculate the specific laeat capacity of cion

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b) A student heats up a 5000 mathrm(~g) block of cion from 18^circ mathrm(C) to 46^circ mathrm(C) , transforming 62,8 mathrm(~kJ) of energy. Calculate the specific laeat capacity of cion

b) A student heats up a 5000 mathrm(~g) block of cion from 18^circ mathrm(C) to 46^circ mathrm(C) , transforming 62,8 mathrm(~kJ) of energy. Calculate the specific laeat capacity of cion

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NeilMaster · Tutor for 5 years

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First, calculate the temperature change:<br /><br />\( \Delta T = T_{f} - T_{i} = 46^{\circ}C - 18^{\circ}C = 28^{\circ}C \)<br /><br />Next, substitute the given values into the specific heat capacity formula:<br /><br />\( cp = \frac{Q}{m \cdot \Delta T} = \frac{62.8 K}{5000 g \cdot 28^{\circ}C} \)<br /><br />Perform the calculation to find the specific heat capacity of iron.

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## Step 1: Understand the Problem<br />The problem is asking us to find the specific heat capacity of iron. The specific heat capacity is the amount of heat energy required to change the temperature of a unit mass of the substance by one degree. It is generally presented in joules(g.Celcius^-1) where 'g' is grams and 'Celcius^-1' is per degree Celsius. <br /><br />## Step 2: Identify the Given Values<br />From the problem, we are given the following values:<br />- The mass of the iron block, \( m = 5000 \) g<br />- The initial temperature, \( T_{i} = 18^{\circ}C \)<br />- The final temperature, \( T_{f} = 46^{\circ}C \)<br />- The energy transferred, \( Q = 62.8 \) K<br /><br />## Step 3: Calculate the Temperature Change<br />The temperature change, \( \Delta T \), is calculated by subtracting the initial temperature from the final temperature. <br /><br />### \( \Delta T = T_{f} - T_{i} \)<br /><br />## Step 4: Use the Specific Heat Capacity Formula<br />The formula for calculating specific heat capacity (cp) is:<br /><br />### \( cp = \frac{Q}{m \cdot \Delta T} \)<br /><br />We can substitute the given values into this formula to find the specific heat capacity.
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