Respuesta :
To solve this question you need to know the specific heat of the object, which is water. The specific heat of water is 1 calorie/gram °C = 4.2joule/gram °C. The energy needed would be:
Q= c*m*ΔT
Q= 1 calorie/gram °C * 250 g * (85-25°C )
Q= 15000 calorie= 63000J
Q= c*m*ΔT
Q= 1 calorie/gram °C * 250 g * (85-25°C )
Q= 15000 calorie= 63000J
Heat energy is given by ; Mcθ
where m is the mass, c is the specific heat capacity and θ is the change in temperature;
Mass of water is 250 g, specific heat of water is 4.18 J/g/k, and θ is the change in temperature
thus; heat = 250 × 4.18 × 65
= 67925 Joules or 67.925 kJ
where m is the mass, c is the specific heat capacity and θ is the change in temperature;
Mass of water is 250 g, specific heat of water is 4.18 J/g/k, and θ is the change in temperature
thus; heat = 250 × 4.18 × 65
= 67925 Joules or 67.925 kJ
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