What is the heat change if 263 g of solid gold increases in temperature by 20 K to 1 d.p?(The molar heat capacity of solid gold is 25.42 J mol-1 K-1)

Sure, to calculate the heat change, we can use the formula:

q = mcΔT

where: q = heat energy m = mass c = specific heat capacity ΔT = change in temperature

However, we are given the molar heat capacity, not the specific heat capacity. The molar heat capacity is the heat capacity per mole of substance. To convert this to specific heat capacity (heat capacity per gram), we need to divide by the molar mass of gold, which is 197.0 g/mol.

So, the specific heat capacity of gold is 25.42 J mol-1 K-1 / 197.0 g/mol = 0.129 J g-1 K-1.

Now we can substitute the values into the formula:

q = (263 g)(0.129 J g-1 K-1)(20 K) = 679.14 J

So, the heat change is approximately 679.1 J to 1 decimal place.

Sure, I can help with that. Here are the steps to calculate the heat change:

1. First, we need to find the number of moles of gold. The molar mass of gold is approximately 197.0 g/mol. So, the number of moles (n) is given by the formula:

   n = mass / molar mass n = 263 g / 197.0 g/mol = 1.34 mol

2. Next, we use the formula for heat change (q), which is given by:

   q = n * C * ΔT

   where: n = number of moles C = molar heat capacity ΔT = change in temperature

3. Substituting the given values into the formula, we get:

   q = 1.34 mol * 25.42 J/mol*K * 20 K = 679.24 J

So, the heat change when 263 g of solid gold increases in temperature by 20 K is approximately 679.2 J to 1 decimal place.