P= VI = 29.6 watts
E= P*t = 14800 j
4.186 J = 1 c
So E= 14800/4.186 cal = 3535.6 cal
Q = c (specific heat) * m * T
T= Q/(m*C) = 3535.6/260* 0.220 = 61.8 C
T = Tf - ti
tf = 61.8 + 24 = 84.8 C
a resistor has a voltage of 6.65 V and a current of 4.45 A. it is placed in a calorimeter containing 200g of water at 24 degree Celsius. the calorimeter is aluminum (specific heat= 0.220 cal/g-Celsius), and its mass is 60g. the heat capacity of the resistor itself is negligible. what is the temperature of the system 500 s later if all the electrical energy goes into heating the water and the calorimeter?
Please show me steps. The lab shows some equations i can use, but I do not fully understad.
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