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Heat Equation (Numerical Problem)

Q.What quantity of heat energy is needed if 20 liters of water is to be heated from 200c to 350c? (Sp.heat capacity of water is 4200J/kg0C) Soln→ Mass of water m = 20 L = 20 kg Initial temp. t 1 = 200c Final temp. t2 = 350c. Sp. heat capacity of water s =4200 J/kg0C Quantity of heat Q = ? By the formula, Q = m s dt = 20 × 4200 × (t2 − t1) = 84000 (35 − 20) = 84000 × 15 =1260000 J = 1260 kJ Q.Calculate the amount of heat required to raise the temperature of 100gm water from 100c to 900c? (Sp.heat capacity of water is 4200J/kg0C) Q.What amount of heat energy is required to raise the temperature of 5 kg copper from 100c to 900c? (Sp.heat capacity of copper is 400J/kg0C)	Q.The sp.heat capacity of mercury is 138 J/kg0C. How much heat is required to increase the temperature of 2 kg of mercury by 500c? Soln→ Mass of mercury m = 2 kg Final change in temp. dt = 500c Sp. heat capacity of water s =138 J/kg0C Quantity of heat Q = ? By the formula, Q = m s dt = 2 × 138 × 50 = 13800 J = $\frac{\mathrm{13800}}{\mathrm{1000}}$ KJ =13.8 KJ Q.Calculate the amount of heat required to raise the temperature of a frying pan by 800c if the mass of the pan is 0.5kg and its specific heat capacity is 480 J/kg0C. Q.Sona needs 20 kg of water at 360c to take a bath, but her water is only at 160c. calculate the minimum amount of heat energy she required to heat the water to the required temperature.
Q.If 15 kilojoules heat energy is required to raise the temperature of a substance of mass 2 kg through 200c, calculate the specific heat capapcity of iron? Soln→ Quantity of heat Q = 15 KJ = 15 × 1000 J = 150000 J Mass of iron m = 2 kg Final change in temp. dt = 200c Sp. heat capacity of water s = ? By the formula, Q = m s dt s = $\frac{Q}{\mathrm{m\; dt}}$ = $\frac{\mathrm{15000}}{\mathrm{2\; \times \; 20}}$ = $\frac{\mathrm{15000}}{\mathrm{40}}$ =375 J/kg0C Q. When 9.4 KJ of heat energy is supplied to 2kg mass of an iron ball the temperature of the ball increases by 100c, calculate the specific heat capapcity of iron. Q. If 2.2 × 105 joule of heat energy is given to 2 kg of water, calculate the increase in temp. Specific heat capapcity of water is 4200j/kg0c.	Q.If a metal ball having of mass 2000 gram and temp. 3000c is dropped in 5 kg water at 200c, the final temperature of water reases 350c. Calculate the sp.heat capacity of the metal ball? Soln→ mass of metal m1 = 2000gm = 2kg Initial temperature t1 = 3000c Final temperature t2 = 350c Sp. heat capacity metal ball S = ? mass of water m2 = 5 kg Initial temperature t1 = 200c Final temperature t2 = 350c Since metal ball is dropped in 5kg water. By the principle of calorimetry Heat lost by metal ball = Heat gain by water or m1 sm dt = m2 sw dt or 2 × sm ( 300 − 35) = 5 × 4200 × (35-20) or 2 × sm × 265 = 21000 × 15 or 530 × sm =315000 or sm =$\frac{\mathrm{315000}}{\mathrm{530}}$ or sm = 594.33 J/kg0. Q.10 kg of water at 50c is mixed with 5 kg water at 900c. Calculate the temperature of mixture. Q. A piece of iron having mass 20 kg and temperature 20000c is dropped into 10kg of water having 250 temp. Calculate the final temp. of the mixture, Sp. heat capacity of iron is 470j/kg0c and Sp. heat capacity of water is 4200J/kg0c
Q. One kg water at 200c is given 84000J heat energy then what will be the final temperature? Soln→ mass of water m = 1kg Initial temperature t1 = 200c Quantity of Heat Q = 84000 J Final temperature t2 ? By the fromula Q = m s dt or, dt = $\frac{Q}{\mathrm{m\; s}}$ or, t2 − t1 = $\frac{\mathrm{84000}}{\mathrm{1\; \times \; 42000}}$ or, t2 − 20 = 2 or, t2 = 2 + 20 t2 = 200 Q. An electric heater of power 1000w gives 3.6 × 106J heat energy in one hour. With all this heat energy 50 kg water initially at 300c heated then what will be the final temperature?	Q. 500 gm mercury is cooled from 100c to − 100c. What amount of heat energy is evolved? (Sp. heat capacity mercury is 138J/kg0c Soln→ mass of mercury m = 500gm = $\frac{\mathrm{500}}{\mathrm{1000}}$=0.5 kg Initial temperature t1 = 100c Final temperature t2 = − 100c Quantity of Heat energy evolved Q = ? By the fromula, Q = m s dt or, Q = 0.5 × 138 (t2 − t1) or, Q = 69 × ( − 10 − 10) or, Q = 69 × ( − 20) or Q = − 1380 J or Q = 1380 J [here, (−)ve sign indicates heat energy evolved.]